Answer:
The answer is C
Explanation:
While playing football, Chris runs 4.5 m at 20⁰ south of west. If an opponent was trying to tackle him how far west and south would he have to run to reach him?
Answer:
South = 1.5m
West =4.2m
Explanation:
Kindly see attached a rough draft of the situation
Step one
Given data
From the sketch the direction of the player is along the resultant of the triangle, corresponding to the Hypotenuse
Step two:
Hence for an opponent to tackle him towards the south, he must be at
sin θ= opp/hyp
sin 20=x/4.5
x=sin 20*4.5
x=0.342*4.5
x= 1.5m
Also, for an opponent to tackle him towards the south, he must be at
cos θ= adj/hyp
cos 20=y/4.5
y=cos 20*4.5
y=0.93*4.5
y= 4.2m
If someone has a curve in his or her spine, what other bones might also be affected by the way the curve shifts that person’s weight?
Answer:
Curve in spine can effect his or her other body parts too in addition to spine. The neck or head can be tilted forward, shoulder could be hunched, hip bone can be uneven, ribs can be effected leading to disorder of lung.
Explanation:
A boy throws a steel ball straight up. Consider the motion of the ball only after it has left the boy's hand but before it touches the ground, and assume that forces exerted by the air are negligible. For these conditions, the force(s) acting on the ball is (are):
a) a downward force of gravity along with a steadily decreasing upward force.
b) a steadily decreasing upward force from the moment it leaves the boy's hand until it reaches its highest point; on the way down there is a steadily increasing downward force of gravity as the object gets closer to the earth.
c) an almost constant downward force of gravity along with an upward force that steadily decreases until the ball reaches its highest point; on the way down there is only a constant downward force of gravity.
d) an almost constant downward force of gravity only.
e) none of the above. The ball falls back to the ground because of its natural tendency to rest on the surface of the earth.
Answer:
c) an almost constant downward force of gravity along with an upward force that steadily decreases until the ball reaches its highest point; on the way down there is only a constant downward force of gravity.
Explanation:
When the boy throws the ball upward, there are two forces acting on the ball. First one is the force of gravity, which is constant and equal tot he weight of the ball, acting down. The second force is the force with which the boy throws the ball. This force acts upward and decreases as the ball moves up. This force becomes zero when the ball reaches highest point.
While, on its way down the ball has only one force acting on it. That force is the constant force of gravity acting downward. This gravity force is equal to the weight of the ball.
Hence, the correct option is:
c) an almost constant downward force of gravity along with an upward force that steadily decreases until the ball reaches its highest point; on the way down there is only a constant downward force of gravity.
The force acting on the ball is an almost constant downward force of gravity along with an upward force that steadily decreases until the ball reaches its highest point; on the way down there is only a constant downward force of gravity. Thus the option C is the correct option.
Given,
Force exerted by the air is negligible.
Now the forces acting on the ball after the realese,
Gravity- This force acts on the body and will remain constant until the ball is in the air. This force is equal to the weight ([tex]W[/tex]) which is equal to the product of the mass [tex]m[/tex] and the gravitational force [tex]g[/tex] of the earth.
[tex]W=mg[/tex]
Force applied by the boy- This is the force by which the boy through the ball. This will act upward and keep on decreasing as the ball moves upward. This force becomes zero when the body reaches the highest point.Now after reaching the highest point, the ball starts to come down. In this case, only one force will act on it which is the gravity force and this is equal to the weight of the body.
Hence, the force acting on the ball is an almost constant downward force of gravity along with an upward force that steadily decreases until the ball reaches its highest point; on the way down there is only a constant downward force of gravity. Thus the option C is the correct option.
For more about the gravity follow the link below-
https://brainly.com/question/1479537
Which is the length of the route between two points
Answer:
Can you show us the route and then I would be happy to help :)
Explanation:
A sky diver jumps out of a plane and reaches terminal velocity at 91 meters per second. How far will she fall in 35 seconds?
Can I get this answer quick!
Explain why people have different skin colors?
I think it because of UV rays ultra violet ray which can make their colors different
A flagpole 24 meters high casts a shadow 33 meters long. What is the distance between the tip of the shadow and flagpole? Round your answer to the nearest meter
Answer:
The distance between the tip of the shadow and flagpole is 40.80 m.
Explanation:
Let the flagpole be represented by AB and the shadow BC . We have to find the distance between C and A .
Considering the right angle CA is the hypotenuse .
Using the Pythagoras theorem
(Hypotenuse)² = (base)² + (perpendicular)²
CA= √(24)² + (33)²
CA= √576+1089
CA= 40.80 m
The distance between the tip of the shadow and flagpole is 40.80 m.
a bus is moving with the velociity of 36 km/hr . after seeing a boy at 20 m ahead on the road, the driver applies the brake and the bus gets stopped at 10 m distance. Now, calculate acceleration as well as time taken by this bus to stop.
Answer:
Assumption: the acceleration of this bus is constant while the brake was applied.
Acceleration of this bus: approximately [tex]\left(-6.0\; \rm m \cdot s^{-2}\right)[/tex].
It took the bus approximately [tex]1.7\;\rm s[/tex] to come to a stop.
Explanation:
Quantities:
Displacement of the bus: [tex]x = 10\; \rm m[/tex].Initial velocity of the bus: [tex]\displaystyle u = 36\; \rm km \cdot hr^{-1} = 36\; \rm km \cdot hr^{-1}\times \frac{1\; \rm m \cdot s^{-1}}{3.6\; \rm km\cdot hr^{-1}} = 10\; \rm m \cdot s^{-1}[/tex].Final velocity of the bus: [tex]v = 0\; \rm m\cdot s^{-1}[/tex] because the bus has come to a stop.Acceleration, [tex]a[/tex]: unknown, but assumed to be a constant.Time taken, [tex]t[/tex]: unknown.Consider the following SUVAT equation:
[tex]\displaystyle x = \frac{1}{2}\, \left(a\, t^2\right) + u\, t[/tex].
On the other hand, assume that the acceleration of this bus is indeed constant. Given the initial and final velocity, the time it took for the bus to stop would be inversely proportional to the acceleration of this bus. That is:
[tex]\displaystyle t = \frac{v - u}{a}[/tex].
Therefore, replace the quantity [tex]t[/tex] with the expression [tex]\displaystyle \left(\frac{v - u}{a}\right)[/tex] in that SUVAT equation:
[tex]\displaystyle x = \frac{1}{2}\, \left(a\, \left(\frac{v -u}{a}\right)^2\right) + u\, \left(\frac{v - u}{a}\right)[/tex].
Simplify this equation:
[tex]\begin{aligned}x &= \frac{1}{2}\, \left(a\, {\left(\frac{v -u}{a}\right)}^2\right) + u\, \left(\frac{v - u}{a}\right) \\ &= \frac{1}{2}\left(\frac{{(v - u)}^2}{a}\right) + \frac{u\, (v - u)}{a} =\frac{1}{a}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right)\end{aligned}[/tex].
Therefore, [tex]\displaystyle a= \frac{1}{x}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right)[/tex].
In this question, the value of [tex]x[/tex], [tex]u[/tex], and [tex]v[/tex] are already known:
[tex]x = 10\; \rm m[/tex].[tex]\displaystyle u =10\; \rm m \cdot s^{-1}[/tex].[tex]v = 0\; \rm m\cdot s^{-1}[/tex].Substitute these quantities into this equation to find the value of [tex]a[/tex]:
[tex]\begin{aligned} a &= \frac{1}{x}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right) \\ &= \frac{1}{10\; \rm m}\times \left(\frac{{\left(0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}\right)}^2}{2} + \left(0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}\right)\times 10\; \rm m \cdot s^{-1}\right)\\ &\approx -6.0\; \rm m \cdot s^{-2}\end{aligned}[/tex].
(The value of acceleration [tex]a[/tex] is less than zero because the velocity of the bus was getting smaller.)
Substitute [tex]a \approx -6.0\; \rm m \cdot s^{-2}[/tex] (alongside [tex]u = 10\; \rm m \cdot s^{-1}[/tex] and [tex]v = 0\; \rm m \cdot s^{-1}[/tex]) to estimate the time required for the bus to come to a stop:
[tex]\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}}{-6.0\; \rm m \cdot s^{-2}} \approx 1.7\; \rm s\end{aligned}[/tex].
which factor is needed to describe velocity but is not needed to describe speed
Answer:
Magnitude and direction describe velocity
However if there is only magnitude its mean it is speed.
Fun : probably you have seen speed movie. It could not have director because if speed is directed then it will be velocity lol..
A cannon is fired from the edge of a cliff, which is 60m above the sea. The cannonball's initial velocity is 88.3m/s and it is fired at an upward angle of 34.5° to the horizontal. Determine: (a) the time the ball is in the air: (b) the impact velocity: (c) the horizontal distance out from the base of the cliff that the ball strikes the water.
Answer:
a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m
Explanation:
a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.
So y - y₀ = ut - 1/2gt²
y - y₀ = (v₀sinθ)t - 1/2gt²
substituting the values of the variables into the equation, we have
0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²
- 60 = 50t - 4.9t²
So, 4.9t² - 50t - 60 = 0
Using the quadratic formula to find t,
[tex]t = \frac{-(-50) +/- \sqrt{(-50)^{2} - 4 X 4.9 X -60} }{2 X 4.9} \\t = \frac{50 +/- \sqrt{2500 + 1176} }{9.8} \\t = \frac{50 +/- \sqrt{3676} }{9.8} \\t = \frac{50 +/- 60.63 }{9.8} \\t = \frac{50 + 60.63 }{9.8} or t = \frac{50 - 60.63 }{9.8} \\t = \frac{110.63 }{9.8} or t = \frac{-10.63 }{9.8} \\t = 11.29 sor -1.085[/tex]
Since t cannot be negative, t = 11.29 s
b. We first need to find the impact vertical velocity component. Using
v = u - gt where u = initial vertical velocity component = v₀sinθ and t = 11.29 s and g = 9.8 m/s². So,
v = v₀sinθ - gt
= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s
= 50.01 m/s - 110.64 m/s
= -60.63 m/s
Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.
The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)
= √((-60.63 m/s)² + (72.77 m/s)²)
= √((3676 m²/s² + 5295.48 m²/s²)
= √(8971.48 m²/s²
= 94.72 m/s
The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°
So the impact velocity is 94.72 m/s at -39.8°
c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m
9. According to statistical theory, 68% of your measurements of time should fall within the range of and of therefore about 3 of your 5 measurements should satisfy this condition. How many of your measurements fall within this range? Are your errors random according to this theory?
Answer:
Yes
Explanation:
Yes, this is a random error generating because of statistical constraint. We only have finite number of data points. As per this, if we plot our observation we will get a gaussian (inverse bell ) shaped curve with mean equal to central value.
Select the correct answer.
What condition should be met for the stellar parallax method to be useful?
A.
the star should be situated within the Milky Way
B.
the star should be brighter than the Sun
C.
the star should be close enough to the Sun to show a measurable parallax
D.
the star should be visible with the naked eye
Answer:
The answer is: C. the star must be close enough to the Sun to show measurable parallax.
Explanation:
In the field of astronomy, parallax is called stellar parallax in distant stars.
The stars for which parallax measurements have been made are those that are relatively close to the sun on the galactic scale.
The answer is: C. the star must be close enough to the Sun to show measurable parallax.
When a bug hits a windshield of a car, the force of the car on the bug equals the force of the bug on the car.
O True
O False
True. The force of the car on the bug will be equal to the force of the bug on the car.
Newton's third lawThis statement is true according to Newton's third law of motion.
According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction.
In this case, the action is the force of the car on the bug, and the reaction is the force of the bug on the car.
Therefore, the force of the car on the bug is equal in magnitude and opposite in direction to the force of the bug on the car.
More on Newton's law can be found here: https://brainly.com/question/15280051
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Two blocks are arranged at the ends of a mass-less string as shown in the figure. The system starts from rest. When the 3.26 kg mass has fallen through 0.386 m, its downward speed is 1.28 m/s. What is the frictional force between the 4.52 kg mass and the table? (round your answer to the 4th decimal place)
Answer: 002 10.0points Two blocks are arranged at the ends of a massless string as shown in . 96 kg mass has fallen through 0 . 415 m, its downward speed is 1 . 25 m / s.
Explanation: I hoped that helped!!
Answer:
2.5 kg
Explanation:
A construction worker accidentally drops a brick from a high scaffold. a. What is the brick's velocity after 4.0 s? b. How far does the brick fall during this time?
Answer:
A. 39.2 m/s
B. 78.4 m
Explanation:
Data obtained from the question include:
Time (t) = 4 s
Acceleration due to gravity (g) = 9.8 m/s²
A. Determination of the brick's velocity.
Time (t) = 4 s
Acceleration due to gravity (g) = 9.8 m/s²
Velocity (v) =?
v = gt
v = 4 × 9.8
v = 39.2 m/s
Thus, the brick's velocity after 4 s is 39.2 m/s
B. Determination of how far the brick fall in 4 s.
Time (t) = 4 s
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) =?
h = ½gt²
h = ½ × 9.8 × 4²
h = 4.9 × 16
h = 78.4 m
Thus, the brick fall 78.4 m during the time.
The velocity after 4 seconds is 40 m/s and the height is 80 m.
From the equations of motion under gravity;
v = u + gt
V = final velocity of the brick
u = initial velocity of the brick
g = acceleration due to gravity = 10 ms-2
t = time taken
Note that, u = 0 m/s because the brick was dropped from a height.
v = 0 + 10(4)
v = 40 m/s
The height is obtained from;
h = ut + 1/2 gt^2
u = 0 m/s because the brick was dropped from a height.
h = 1/2 gt^2
h = 1/2 × 10 × (4)^2
h = 80 m
Learn more: https://brainly.com/question/2375972
(NEED HELP ASAP)
In a lab investigation one group of students (group a) measures the speed of a 0.1 toy car to be 2.5 m/s at the bottom of a ramp another group of students (group b) measures the speed of the car to be 3 m/s at the bottom of the ramp the cars starting position at the top of the ramp is 1 m high.
a. what is the potential energy of the car at the beginning of the experiment before it's speed is measured?
b. Calculate the kinetic energy of the car for group a.
C. Calculate the kinetic energy for the car for group b.
Answer:
a = 0.98J
Explanation:
PE = mgh
PE = 0.1 x 9.8 x 1
PE = 0.98J
(a) The potential energy of the car at the beginning of the experiment is 9.8(m) J.
(b) The kinetic energy of the car for group a is 3.125(m) J.
(c) The kinetic energy of the car for group b is 4.5(m) J.
Potential energy of the car at the beginning of the experimentThe potential energy of the car is calculated as follows;
P.E = mgh
where;
m is mass of the carg is accleration due to gravityh is height of the carP.E = m x 9.8 x 1
P.E = 9.8(m) J
Kinetic energy of the car for group aK.E = ¹/₂mv²
K.E = ¹/₂ x m x (2.5)²
K.E = 3.125(m) J
Kinetic energy of the car for group bK.E = ¹/₂mv²
K.E = ¹/₂ x m x (3)²
K.E = 4.5(m) J
Learn more about kinetic energy here:https://brainly.com/question/25959744
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Which combustion product is a deadly gas that you cannot see or smell? O sulfur dioxide O carbon monoxide O carbon dioxide O sulfur monoxide
Answer:
carbon monoxide
Explanation:
* 1 po
What are the health benefits of having muscular strength and endurance?
Answer:
i can`t help without showing me the choses
Explanation:
Lily can run 100 m in 10s. How fast was her average speed?
Answer:
The answer is 10 m/sExplanation:
The average speed can be found by using the formula
[tex]v = \frac{d}{t} \\ [/tex]
d is the distance
t is the time taken
From the question we have
[tex]v = \frac{100}{10} \\ [/tex]
We have the final answer as
10 m/sHope this helps you
A researcher may use a two-tailed test to evaluate a directional hypothesis.
Please select the best answer from the choices provided
OT
OF
Answer:
true
Explanation:
just took the test
Answer:
True
Explanation:
if a truck goes 30 kilometers in 30 minutes what is the average speed
Answer:
60 km per hour
Explanation:
if you drive for 30 min and go 30km then if you drive for 60 min(1hr) then you would have driven 60km.
Answer:
60
Explanation:
hope this helps
Can y’all please help
Answer:
The cat's speed when it slid off the table was 2.53 m/s
Explanation:
Horizontal Motion
When an object is thrown horizontally at a speed v, from a height h, it describes a curved path ruled by gravity, until it hits the ground.
the range or maximum horizontal distance traveled by the object can be calculated as follows:
[tex]\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}[/tex]
The situation described in the problem fits into the concept of a horizontal launch when the cat slides off the table and gets to the ground some distance ahead.
It's given the range as d=1.3m and the height h=1.3 m, thus we can find the cat's speed by solving the equation for v:
[tex]\displaystyle v=d\cdot\sqrt{\frac {g}{2h}}[/tex]
Where [tex]g=9.81\ m/s^2[/tex]
Substituting values:
[tex]\displaystyle v=1.3\cdot\sqrt{\frac {9.81}{2\cdot 1.3}}[/tex]
[tex]\displaystyle v=1.3\cdot\sqrt{\frac {9.81}{2.6}}[/tex]
Calculating:
v = 2.53 m/s
The cat's speed when it slid off the table was 2.53 m/s
What will happen if the atom rearrange?
1. a new substance will form
2. start to boil
3. nothing will happen
choose the correct answer
Answer:
In a chemical reaction, only the atoms present in the reactants can end up in the products. No new atoms are created, and no atoms are destroyed. In a chemical reaction, reactants contact each other, bonds between atoms in the reactants are broken, and atoms rearrange amd form new bonds to make the products.
8
If you put 100 carbon atoms in a box. How many carbon atoms remain after 2 half-lives or
11460 years? (you might need to review half-lives to answer this question) *
(1 Point)
100 atoms
50 atoms
25 atoms
12 atoms
Answer:
25 atoms
Explanation:
A small plane flew 200 mph with a 40 mph tailwind for 480 miles. The plane returned home at a speed of 200 mph with a 40 mph headwind for 480 miles. How much longer did the flight home take?
Answer:
1 hour
Explanation:
The tailwind usually adds to the speed of airplanes while the headwind normally decreases the speed of airplanes.
For the initial journey of 200 mph with a 40 mph tailwind speed for 480 miles:
Total speed = 200 + 40 = 240 mph
Also,
Speed = distance/time
time = 480/240 = 2 hours
For the return journey of 200 mph with a 40 mph headwind for 480 miles:
Total speed = 200 - 40 = 160 mph
time = 480/160 = 3 hours
The initial flight took 2 hours while the flight back home took 3 hours. Hence it took 1 hour longer to get back home
For brainliest !!!!!!! Do it
Answer and Explanation:
Using a = F/m, we get:
Acceleration for Road A = 480/62 = ~7.74 m/s^2
Acceleration for Road B = 240/62 = ~3.87 m/s^2
Acceleration for Road C = 600/62 = ~9.68 m/s^2
If the net force becomes higher than 600N and the mass stays constant, the acceleration will increase proportionally to how much the force increases and the acceleration will be greater than the acceleration when the force is 600N.
I hope this helps! :)
17.
A car is accelerated at 4.0 m/s2 from rest. How long does it take the car to reach a speed of 28 meters per second?
Answer:
t = 7 seconds
Explanation:
Given that,
The acceleration of a car, a = 4 m/s²
The initial velocity of the car, u = 0
Final velocity of the car, v = 28 m/s
We need to find how much time the car to reach a speed of 28 m/s. Using the definition of the acceleration of a car. It is given by :
[tex]a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{28}{4}\\\\t=7\ s[/tex]
So, it will take 7 seconds to reach a speed of 28 m/s.
18) Why do you think divergent plate boundaries and transform fault boundaries only occur at places where the crust is the same on both plates?
Answer:
A divergent boundary occurs when two tectonic plates move away from each other. Along these boundaries, earthquakes are common and magma (molten rock) rises from the Earth's mantle to the surface, solidifying to create new oceanic crust. When two plates come together, it is known as a convergent boundary.
Explanation:
please help me.
During a football game, one of the players on the home team kicks the football that has a
mass of 0.6 kg so that the ball accelerates toward the opposing team at
23 m/s2. If no other forces act on the ball, how much force did the kicker apply to the
football?
Answer:
[tex]Force = 13.8N[/tex]
Explanation:
Given
[tex]Mass = 0.6kg[/tex]
[tex]Acceleration = 23m/s^2[/tex]
Required
Determine the applied force
From the question, we understand that no other force acts on the ball.
i.e. the only applied force on the ball is the force applied by the striker.
So, we apply Newton's second law to solve this question.
And this implies that:
[tex]Force = Mass * Acceleration[/tex]
[tex]Force = 0.6kg * 23m/s^2[/tex]
[tex]Force = 13.8N[/tex]
Hence, the applied force by the striker on the ball is 13.8N
the photos below show four pairs of objects. which pair of objects is experiencing the greatest gravitational force.A. the person and dog B. The person and the ball C. the person and earth D. the person and the building
Answer:
C.
Explanation:
the earth has a high greavitational pull
Answer: C the person and earth
Explanation: I just answered it on a p e x