Answer:
yes
Explanation:
people still skateboard that is an easy question
What is the most frequent method used to slow a vehicle that is done by releasing pedal pressure
trail braking
controlled braking
threshold braking
coasting
Answer: Trail Braking!
Explanation: Trail braking is used to smoothly and gradually reduce brake pedal pressure at the end of a braking maneuver.
Option A. trail braking.
What is trail braking and how do you do it?In four-wheel vehicles, trail braking is using the brakes past the corner entrance, as opposed to the normally taught practice of releasing the brakes before starting the turn. It creates weight transfer to the front tires, increasing their traction and reducing understeer.
What is the purpose of trail braking?Trail-braking allows a driver to brake later by extending the braking zone into the corner. It also improves the car's turn-in response by increasing the load on the front tires to maximize grip. Together, these result in higher corner entry speed and reduced time through the corner.
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A 12-V battery is connected across a 100-Ω resistor. How many electrons flow through the wire in 1.0 min?
Answer:
The quantity of charge or electron flowing the wire in the given time is 4.5 x 10¹⁹ electrons.
Explanation:
Given;
emf of the battery, V = 12 V
resistance of the resistor, R = 100-Ω
time of current flow, t = 1 min
charge of 1 electron = 1.602 x 10¹⁹ C
The current through this circuit is given by;
I = V / R
I = (12) / (100)
I = 0.12 A
The quantity of charge or electron flowing the wire in the given time is calculated as;
Q =It
where;
I is the current flowing through the wire
t is the time of current flow = 1 x 60s = 60 s
Q = 0.12 x 60
Q = 7.2 C
1.602 x 10⁻¹⁹ C --------------- 1 electron
7.2 C -----------------------------? electron
[tex]= \frac{7.2 }{1.602*10^{-19}} \\\\= 4.5*10^{19} \ electrons[/tex]
Therefore, the quantity of charge or electron flowing the wire in the given time is 4.5 x 10¹⁹ electrons.
Suppose the electric field between two points separated by 4 meters is 37 Volt/m. What is the electric potential (in Volt) between the two points? Use exact numbers; do not estimate.
Answer:
148 V
Explanation:
The electric field can be obtained using the formula;
E= V/d
where;
E is the electric field intensity = 37 volt/m
V is the potential difference (the unknown)
m is the distance of separation = 4m
V = Ed
V = 37 volt/m * 4 m
V= 148 V
Question 5 of 5
What is a definite sign of overtraining?
O A. Depression
B. Declining athletic performance
O C. A ravenous appetite
O D. Big bursts of energy
( the answer is B. i just put it on here because i didn’t see it and i got it correct :)
what causes the coriolis effect
A. Earths orbit around the sun.
B. Wind currents.
C. Earths rotation around its axis
D. Uneven solar heating of earth
plzzzzzzzzzzzzzzzzzzzzzzzzzz help 20 points
Answer:
1.23
Explanation:
[tex]{\underline{\pink{\textsf{\textbf{ Answer : }}}}}[/tex]
➩ 1.23 feet
[tex]{\underline{\purple{\textsf{\textbf{Explanation : }}}}}[/tex]
Given :
Simon cuts a pipe that was 4.92 feet long Then he cuts it into four equal pieces.To find :
What is the length of the each piece.Solution :
As it is told that it's divided into four equal pieces
Therefore,
We must divide it by 4 to get the length of each piece.
So,
[tex] \sf \to \: \frac{4.92}{4} \\ \sf \to \: 1.23 \: feet \: ans.[/tex]
A body of moment of inertia I=0.80kgm2 about a fixed axis, rotating with constant angular velocity of 100 rad s-1, then torque acting on it will be:
a)80Nm
b)zero
c)160Nm
d)120Nm
A centripetal force of 210 N acts on a 1,600-kg satellite moving with a speed of 5,500 m/s in a circular orbit around a planet. What is the radius of its orbit?
Answer:
230476.19kmExplanation:
Step one:
given
Force F= 210N
mass m= 1600kg
velocity v=5500m/s
Step two
Required is the radius r
the expression for the force is
[tex]F_c = \frac{mv^2}{r}[/tex]
substitute
210=1600*5500^2/r
cross multiply we have
210r=48400000000
divide both side by 210
r=230476190.476m
r=230476.19km
If you were standing at the center of a circular wave what would you see in all directions?
a.waves moving away from you
b.waves moving towards you
c.waves moving across you
d.no movement at all
which statement accurately describes the relationship between force and momentum?
A. As the mass of an object increases its momentum increases, and it takes more force to change its motion.
B. As the velocity of an object increases, its momentum decreases and it takes less force to change its motion.
C. As the mass of an object increases its momentum decreases and it takes less force to change it motion
D. As the velocity of an object decreases its momentum increases and it takes more force to change its motion
Answer:
A
Explanation:
just did it
As the mass of an object increases its momentum increases, and it takes more force to change its motion. So, option A.
What is meant by momentum ?Mass in motion is quantified by momentum, which is the measure of the amount of mass in motion.
Here,
Momentum of an object, which is under motion can be defined as the product of the mass and velocity of the object.
Momentum, P = mv
According to Newton's second law, the force is defined as the rate of change of momentum, or the momentum per unit time.
F = dP/dt
So, force is proportional to the amount of momentum imparted on the object.
Therefore, if the mass or velocity of the object increases, it will eventually cause the momentum to be increased and as a result, the force required to exert on the object will increase.
Hence,
As the mass of an object increases its momentum increases, and it takes more force to change its motion. So, option A.
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A box of books weighing 319 N is shoved across the floor by a force of 485 N exerted downward at an angle of 35 degres below the horizontal.a) If the coeficent of friction between the box and the floor is 0.57, how long does it take to move the box 4 meters, starting from rest?b) If If the coeficent of friction between the box and the floor is 0.75, how long does it take to move the box 4 meters, starting from rest?
Let w, n, p, and f denote the magnitudes of the 4 forces acting on the box.
• w = weight = 319 N
• n = normal force
• p = pushing force = 485 N
• f = friction = µ n, where µ is the coefficient of static friction
The net force on the box points in the direction that the box moves, which is to the right. In particular, this means the box is vertically in equilibrium. Split up the vectors into their vertical and horizontal components, and apply Newton's second law. (I take up and right to be the positive vertical and horizontal directions, respectively.)
• vertical:
p sin(-35°) + n - w = 0
and solving for n,
- (485 N) sin(35°) + n - 319 N = 0
n ≈ 597 N
• horizontal:
p cos(-35°) - f = m a
where a is the magnitude of the net acceleration on the box. Solve for a. Since f = µ n and m = w / g (where g = 9.80 m/s² is the mag. of the acc. due to gravity) we get
p cos(35°) - µ n = (w / g) a
(485 N) cos(35°) - µ (597 N) = (319 N) / (9.80 m/s²) a
a ≈ (12.2 - 18.3 µ) m/s²
(a) If µ = 0.57, then the net acceleration on the box is
a ≈ (12.2 - 18.3 • 0.57) m/s² ≈ 1.75 m/s²
so that the time t required to move the box 4 m is
4 m = 1/2 a t ²
t ≈ √((8 m) / (1.75 m/s²))
t ≈ 2.14 s
(b) The box does not move.
If µ = 0.75, then
a = (12.2 - 18.3 • 0.75) m/s² ≈ -1.55 m/s²
but a negative acc. here means the applied acc. points *opposite* the direction of movement, thus making the box move backward which doesn't make sense. The coefficient of friction is too large for the given applied force to get the box moving. With µ = 0.75, the frictional force to overcome has mag. f ≈ 448 N. But the given push contributes a horizontal force of (485 N) cos(-35°) ≈ 397 N. This mag. needs to be increased in order to get the box moving.
(a) The time taken to move the box 4 meters is 2.14 s.
(b) The box will decelerate when the coefficient of friction is 0.75 and cannot be moved to 4 meters forward.
The given parameters;
weight of the book, W = 319 Napplied force, F = 485 Nangle of inclination, θ = 35 ⁰The mass of the books is calculated as;
[tex]m = \frac{W}{g} \\\\m = \frac{319}{9.8} \\\\m = 32.55 \ kg[/tex]
The normal force on the box is calculated as follows;
[tex]F_n = -W - Fsin\theta\\\\F_n = -319 - (485\times sin35)\\\\F_n = -597.18 \ N[/tex]
The frictional force when the coefficient of friction is 0.57;
[tex]F_f = \mu F_n\\\\F_f = 0.57 \times -597.18\\\\F_f = -340.39 \ N[/tex]
The acceleration of the box is calculated as follows;
[tex]F cos \theta - F_f = ma\\\\(485)cos(35) \ - 340.39 = 32.55 a\\\\56.899 = 32.55a\\\\a = \frac{56.899}{32.55} \\\\a = 1.75 \ m/s^2[/tex]
The time taken to move the box 4 meters is calculated as;
[tex]s = v_0t + \frac{1}{2} at^2\\\\s = 0 + \frac{1}{2} at^2\\\\t = \sqrt{\frac{2s}{a} } \\\\t = \sqrt{\frac{2\times 4}{1.75} } \\\\t = 2.14 \ s[/tex]
(b) The frictional force when the coefficient of friction is 0.75;
[tex]F_f = \mu F_n\\\\F_f = 0.75 \times -597.18\\\\F_f = -447.885 \ N[/tex]
The acceleration of the box is calculated as follows;
[tex]F cos \theta - F_f = ma\\\\(485)cos(35) \ -447.885 = 32.55 a\\\\-50.596 = 32.55a\\\\a = \frac{-50.596}{32.55} \\\\a = -1.55\ m/s^2[/tex]
Thus, the box will decelerate when the coefficient of friction is 0.75 and cannot be moved to 4 meters forward.
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A distant galaxy has a redshift z = 5.82 and a recessional velocity vr = 287,000 km/s (about 96% of the speed of light.) Notice that the equation z=vrcz=vrc does not hold true for recessional velocities approaching the speed of light. What is the distance to the galaxy in light years?
Answer: 4100 Mpc
Explanation:
Since H o = 70 km/s/Mpc
Redshift z = 5.82
Recessional velocity vr = 287,000 km/s
Then, the distance to the galaxy in light years will be:
= Recessional velocity / H o
= 287000 / 70
= 4100 Mpc
A plane is flying due west at 34 m/s. It encounters a wind blowing at 19 m/s south. Find the resultant veloci
Answer:
The resultant velocity has a magnitude of 38.95 m/s
Explanation:
Vector Addition
Given two vectors defined as:
[tex]\vec v_1=(x_1,y_1)[/tex]
[tex]\vec v_2=(x_2,y_2)[/tex]
The sum of the vectors is:
[tex]\vec v=(x_1+x_2,y_1+y_2)[/tex]
The magnitude of a vector can be calculated by
[tex]d=\sqrt{x^2+y^2}[/tex]
Where x and y are the rectangular components of the vector.
We have a plane flying due west at 34 m/s. Its velocity vector is:
[tex]\vec v_1=(-34,0)[/tex]
The wind blows at 19 m/s south, thus:
[tex]\vec v_2=(0,-19)[/tex]
The sum of both velocities gives the resultant velocity:
[tex]\vec v =(-34,-19)[/tex]
The magnitude of this velocity is:
[tex]d=\sqrt{(-34)^2+(-19)^2}[/tex]
[tex]d=\sqrt{1156+361}=\sqrt{1517}[/tex]
d = 38.95 m/s
The resultant velocity has a magnitude of 38.95 m/s
A circular conducting loop with a radius of 1.00 m and a small gap filled with a 10.0 Ω resistor is oriented in the xy-plane. If a magnetic field of 2.0 T, making an angle of 30º with the z-axis, increases to 11.0 T, in 2.5 s, what is the magnitude of the current that will be caused to flow in the conductor?
Answer:
ill get back to this question once i find the answer to it
_______mirrors or lenses always produce smaller images.
Answer:
The images produced by a convex mirror are smaller than the object it reflects. The image produced by a concave lens is always virtual. The image in a convex mirror is always upright and is smaller than the object.
Explanation:
:)
Earth is about 93 million miles from the Sun but only 239,000 miles from the Moon. Likewise, the Sun has a mass that is over 20 million times that of the Moon. Earth experiences a strong gravitational attraction to both of these bodies, but for two different reasons. Using what you know about the factors that influence gravity, what specific characteristic or characteristics contribute to the attraction between Earth and the Moon? What specific characteristic or characteristics contribute to the attraction between Earth and the Sun?
Answer:
Gravity is dependent on the mass of two bodies and the distance between them. There is a strong gravitational attraction between Earth and the Moon because they’re relatively close to one another. There is a strong gravitational attraction between Earth and the Sun because the Sun is so massive
Answer:
Gravity is determined by the mass of two bodies and their separation. Because the Earth and the Moon are so close to one another, they have a tremendous gravitational attraction. Because the Sun is so large, there is a significant gravitational force between Earth and it.
Explanation:
A velocity vector has magnitude 100.0 m/s and make an angle of 160o with the positive x-axis. Determine the x- and y-components of the vector.
Answer:
Vₓ = -93.96 m/s
Vy = 34.2 m/s
Explanation:
The x-component of the velocity vector can be given as follows:
Vₓ = V Cos θ
where,
Vₓ = x-component of velocity vector = ?
V = Magnitude of Velocity Vector = 100 m/s
θ = Angle with positive x-axis = 160°
Therefore,
Vₓ = (100 m/s)Cos 160°
Vₓ = -93.96 m/s
The y-component of the velocity vector can be given as follows:
Vy = V Sin θ
where,
Vy = y-component of velocity vector = ?
V = Magnitude of Velocity Vector = 100 m/s
θ = Angle with positive x-axis = 160°
Therefore,
Vy = (100 m/s)Sin 160°
Vy = 34.2 m/s
Calculate the effective value of g , the acceleration of gravity, at 6800 km , above the Earth's surface.
An electric circuit, powered by a voltage of 100 V, absorbs a power of 4.41 kW; calculate the intensity of the current that runs through the circuit and the energy absorbed by it in 2 hours and 15 minute
Answer:
I = 44.1[amp]
P = 9.92 [kw-hr]
Explanation:
We know that electrical power can be calculated by means of the following expression.
[tex]P=V*I[/tex]
where:
P = power = 4.41 [kW] = 4410 [w]
V = 100 [V]
I = current [amp]
Now we have:
[tex]I=P/V\\I = 4410/100\\I = 44.1 [amp][/tex]
Now the power absorbed during this time is equal to the product of power by time.
[tex]15[min]*\frac{1hr}{60min} =0.25 [hr][/tex]
t = time = 2.25 [hr]
[tex]P_{ab}=4.41*2.25\\P=9.92[kW-hr][/tex]
A cheetah can maintain a maximum constant velocity of 34.2 m/s for 8.70 s. What is
the displacement the cheetah covered at that velocity?
Answer:
297.54mExplanation:
step one:
given data
velocity v=34.2m/s
time t= 8.7s
Step two
Required is the distance the cheetah has covered on the condition
we know that speed= distance/time
make distance subject of formula we have
distance= velocity *time
distance= 34.2*8.7
distance = 297.54m
Therefore the displacement the cheetah covered at that velocity
is 297.54m
What causes rain?
a.air becomes filled with water vapor
b.water vapor condenses on dust particles
c. dust particles can no longer support water droplets
Answer:
the awnser is A
Explanation:
Brainliest Please
Answer:
A
Explanation
Within a cloud, water droplets condense onto one another so its causing so its causing the droplets to grow when they have grown to heavy the fall causing rain.
Suppose a popular FM radio station broadcasts radio waves with a frequency of 96. MHz. Calculate the wavelength of these radio waves. Be sure your answer has the correct number of significant digits.
Answer:
3.125 meters.
Explanation:
(3.0*10^8)/(96*10^6)
= 3.125 meters.
Hope this helped!
A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s.
Complete Question
A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?
Answer:
The position of the object at t = 10s is [tex]X = 38.3 \ m[/tex]
Explanation:
From the question we are told that
The acceleration along the x axis is [tex]a_{x}t = -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)[/tex]
The position of the object at t = 0 is x = -14.0 m
The velocity at t = 0 s is [tex]v_{0}x = 7.10 m/s[/tex]
Generally from the equation for acceleration along x axis we have that
[tex]a_x = \frac{dV_{x}}{dt} = -0.032 (15- t)[/tex]
=> [tex]\int\limits {dV_{x}} \, = \int\limits {-0.032(15- t)} \, dt[/tex]
=> [tex]V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]
At t =0 s and [tex]v_{0}x = 7.10 m/s[/tex]
=> [tex]7.10 = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1[/tex]
=> [tex]K_1 = 7.10[/tex]
So
[tex]\frac{dX}{dt} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]
=> [tex]\int\limits dX = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}[/tex]
=> [tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ K_1t +K_2[/tex]
At t =0 s and x = -14.0 m
[tex]-14 = -0.032 [ 15\frac{0^2}{2} - \frac{0^3 }{6} ]+ K_1(0) +K_2[/tex]
=> [tex]K_2 = -14[/tex]
So
[tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ 7.10 t -14[/tex]
At t = 10.0 s
[tex]X = -0.032 [ 15\frac{10^2}{2} - \frac{10^3 }{6} ]+ 7.10 (10) -14[/tex]
=> [tex]X = 38.3 \ m[/tex]
If you start at a speed of 4m/s and slow down to 2m/s in 4s what is your
acceleration?
Answer:
penis
Explanation:
A school track team member ran for a total of 15.85 miles in practice over 5 days. How many miles did he average per day?
[tex]{\underline{\pink{\textsf{\textbf{ Answer : }}}}}[/tex]
➡ 15.85/5
➡ 3.17 ans.
Which phrase desenbes an irregular galaxy ?
has a round shape
contains many young stars
has arms that extend from the center
Is larger than other types of galaxies
Answer:
contains many young stars
Explanation:
Irregular galaxies have no definite shape, which means that the first option is incorrect. They are definitely not round.
However, they contain many young stars because the degree of star formation is fast. They also contain old stars. Thus, the second choice is correct.
The "spiral galaxy" is the type of galaxy that has arms that extend from the center. These arms look "spiral," which influenced its name. This makes the last choice incorrect.
They are actually smaller than the other types of galaxies. This makes them prone to collisions. This makes the last choice incorrect.
Answer:
Contains many young stars
Explanation:
A tennis ball moves 18 meters northward, then 22 meters
southward, then 14 meters northward, and finally 28 meters
southward.
Answer:
The distance moved is 82 m.
The displacement is 18 m to the south.
Explanation:
The distance is a measure of the total length traveled along the path, while the displacement only takes into account the length between the starting position (departure) and final position (arrival). That is, distance refers to how much space an object travels during its movement, being the amount moved, while displacement refers to the distance and direction of the final position with respect to the initial position of an object.
So, the distance being the sum of the distances traveled, you get:
18 m + 22 m + 14 m + 28 m= 82 m
The distance moved is 82 m.
You know that the tennis ball moves 18 meters to the north, then 22 meters to the south, then 14 meters to the north, and finally 28 meters to the south. Then the tennis ball moves:
northward: 18 m + 14 m= 32 mto the south: 22 m + 28 m= 50 mCalculating the displacement as the difference between the final position and the initial position, you get:
displacement= 50 m - 32 m= 18 m
The displacement is 18 m to the south.
List Five examples from daily life in which you see periodic motion caused by a pendulum
(Marking Brainliest)
Answer:
by a rocking chair, a bouncing ball, a vibrating tuning fork, a swing in motion, the Earth in its orbit around the Sun, and a water wave.
Explanation:
A 50kg boy stands on rough horizontal ground. The coefficient
of static friction, us, is 0.68. The maximum static friction
between the boy and the ground is __N.
Given :
A 50 kg boy stands on rough horizontal ground. The coefficient of static friction, us, is 0.68.
To Find :
The maximum static friction between the boy and the ground is _ N.
Solution :
We know maximum static friction is given by :
[tex]F = \mu mg \\\\F= 0.68\times 50\times 9.8\\\\F = 333.2\ N[/tex]
Therefore, maximum static friction is 333.2 N.
Hence, this is the required solution.
A stone is dropped from a bridge 45 meters above the surface of a river. What is the time required for the stone to reach the water’s surface?
Answer:
C. 3.0 s
Explanation:
Well, I've taken this question before, so I went back and this was the answer I put, which was correct.
Hope this helps! <3 May I have brainliest if it helps you enough?
It will take 3 s for the stone to get to the surface of the water.
We'll begin by listing out what was given from the question. This includes:
Height (h) = 45 m
Time (t) =?NOTE: Acceleration due to gravity (g) is 10 m/s²
From the data above, we can obtain the time taken for the stone to get to the surface of the water as follow:
H = ½gt²45 = ½ × 10 × t²
45 = 5 × t²
Divide both side by 5[tex]t^{2} = \frac{45}{5} \\\\t^{2} = 9[/tex]
Take the square root of both side[tex]t = \sqrt{9}[/tex]
t = 3 sTherefore, it will take 3 s for the stone to get to the surface of the water.
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