A 68 kg deer has a momentum of 952 kg∙m/s. What is its velocity?

Explanation:

In this problem, we are meant to slove for the resultant and the direction of the the vectors given

Given data

let the sail to a point due north be y= 116km

and the point due east be x= 121 km

(a) How far (in km)

The resultant between the two points is the distance between them

[tex]r=\sqrt{x^2+y^2} \\\\r=\sqrt{121^2+116^2} \\\\r=\sqrt{14641+13456} \\\\r=\sqrt{28097} \\\\r=167.62[/tex]

The distance between the points is 167.62

(b) in what direction (as an angle from due east, where north of east is a positive angle) must it now sail to reach its original destination

the direction can be gotten using

tan∅= y/x

∅= tan-1 (y/x)

∅= tan-1(116/121)

∅= tan-1(0.958)

∅= 43.77°

The direction is 43.77°

Answer:

The correct option is a

Explanation:

This question seeks to test a general rule in physics (on charges) which states that like charges repel but unlike charges attract. This means that, a negatively charged substance will repel or not attract another negatively charged material and the same applies to a positively charged substance also. However, a negatively charged substance will attract a positively charged material and vice versa, hence only a negatively charged rod will attract a positively charged hanging ball.

Answer:

Displacement vector represents the average velocity of the bird.

Explanation:

Given that,

A bird flutters around in a tree in a path described by the dark line.

Suppose, given vectors

(a). Time, (b). Displacement, (c). speed, (d). distance

We know that,

Vector quantity :

Vector quantity has direction and magnitude.

Average velocity :

Average velocity is equal to the displacement divided by time.

In mathematically,

[tex]v=\dfrac{D}{t}[/tex]

Where, D = displacement

t = time

We need to find the vector which is represents the average velocity of the bird

Using given data

Average velocity of the bird shows the displacement over the time.

Displacement is the vector quantity.

Hence, Displacement vector represents the average velocity of the bird.

Answer:

The magnitude of the acceleration of the car after the brakes failed is 4 m/s²

Explanation:

The car was originally traveling at 30.0 m/s, that is

The initial velocity, [tex]u[/tex] = 30.0 m/s

The time spent while the car manages to brake is 5.0 seconds, that is

time, [tex]t[/tex] = 5.0 secs

and the distance traveled during this time is

distance, [tex]s[/tex] = 125 m

From one of the equations of kinematics for linear motion,

[tex]s = ut + \frac{1}{2}at^{2} \\[/tex]

Where [tex]a[/tex] is the acceleration

We can determine the deceleration of the car during the first 5.0 seconds

Hence,

From,

[tex]s = ut + \frac{1}{2}at^{2} \\[/tex]

[tex]125 = 30.0(5.0) + \frac{1}{2}(a)(5.0)^{2}[/tex]

[tex]125 =150.0 + 12.5a[/tex]

[tex]12.5a = 125 - 150.0[/tex]

[tex]12.5a = -25\\a = \frac{-25}{12.5}\[/tex]

[tex]a = - 2.0 m/s^{2}[/tex] (Negative sign indicates deceleration)

Now we will calculate the final velocity reached at this time

From,

[tex]v^{2} = u^{2} + 2as[/tex]

Where [tex]v[/tex] is the final velocity

[tex]v^{2} = 30.0^{2} + 2(-2.0)(125)\\v^{2} = 400\\v = \sqrt{400} \\v = 20 m/s \\[/tex]

This is the final velocity reached by the car during the first 5.0 seconds

Now, for the magnitude of the acceleration of the car after the brakes failed,

After the brakes failed,

it travels an additional 150 m further down the road, that is

s = 150m

an additional 5.0 seconds, that is

t = 5.0 seconds

Also, from

[tex]s = ut + \frac{1}{2}at^{2} \\[/tex]

The initial velocity here will be the final velocity for the first 5.0 seconds, that is,

u = 20 m/s

Hence,

[tex]s = ut + \frac{1}{2}at^{2} \\[/tex] becomes

[tex]150 = 20(5.0) + \frac{1}{2}(a)(5.0)^{2}[/tex]

[tex]150 = 100 + 12.5a\\12.5a = 150 - 100\\12.5a = 50\\a = \frac{50}{12.5} \\a = 4m/s^{2}[/tex]

Hence, the magnitude of the acceleration of the car after the brakes failed is 4 m/s²

Answer:

A) Total Distance = 2000 m and Total displacement = 0 m

B) Average Speed = 44.44 m/min and Average Velocity = 0 m/min

C) Average Speed = 0.7407 m/s and Average velocity = 0 m/s

Explanation:

A) Distance to reach ice cream shop from home = 1000 meters

Therefore distance to get back home would also be 1000 meters.

Total distance traveled = 1000 + 1000 = 2000 metres

Since journey started at home and ended at home, then total displacement = 0 metres.

B) Average speed = Total distance/total time.

Total time = 10 + 15 + 20 = 45 minutes

Since total distance = 2000 m

Then;

Average speed = 2000/45

Average speed = 44.44 m/min

Average velocity = Total displacement/total time

Average velocity = 0/45 = 0 m/min

C) We now want answers in B to be in m/s.

Total time = 45 minutes.

From conversion, 60 seconds make 1 minute. Thus, 45 minutes = 45 × 60 = 2700 seconds

Thus;

Average Speed = 2000/2700

Average Speed = 0.7407 m/s

Average displacement = 0/2700 = 0 m/s

Answer:

An object opposes any change in its velocity, either to its direction or to its speed.

Explanation:

Edmentum Answer

Answer:

A calorimeter is can be used to measure the amount of heat released or involved in a chemical reaction. Whereas thermometer can only measures temperature or hotness of a substance. It cannot be used to measure the thermal rate or amount of heat energy of a reaction.

Explanation:

Answer:

Its according from what it is made with if wax it takes longer but if with a craft paper it takes lesser time if paper approximately 20 to 25 minutes

Answer:

Calorimeter is used to measure heat in and represents that in units of joules per kelvin units J/˚C or kJ/K

A calorimeter is can be used to measure the amount of heat released or involved in a chemical reaction.

Whereas thermometer can only measures temperature or hotness of a substance. It cannot be used to measure the thermal rate or amount of heat energy of a reaction. Unit measurement used by thermometer is Celsius (°C).

Explanation:

Recall the formulas,

[tex]x_f=x_i+v_it+\dfrac12at^2[/tex]

[tex]v_f=v_i+at[/tex]

(a)

[tex]10.0\,\mathrm m=\left(19.0\dfrac{\rm m}{\rm s}\right)+\dfrac12\left(-3.50\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]

[tex]\implies \boxed{t\approx0.555\,\mathrm s}[/tex]

(b)

[tex]v_f=19.0\dfrac{\rm m}{\rm s}+\left(-3.50\dfrac{\rm m}{\mathrm s^2}\right)(0.555\,\mathrm s)[/tex]

[tex]\implies \boxed{v_f\approx17.1\dfrac{\rm m}{\rm s}}[/tex]

Answer:

It's increasing with time

Answer:

A. To find height of collision let's find

Speed of collision first

Using

Vf= √2g(h-g)

But for Harvey's ball we have

V²-2gy= 0

y=( v²/2g) so this is height of collision

B.

To find relative speed of the ball

Using V at y to find Vsr

Vs= √2g(h-v²/2g)

= √2gh- v²

Answer:Newton’s law also states that larger bodies with heavier masses exert more gravitational pull, which is why those who walked on the much smaller moon experienced a sense of weightlessness, as it had a smaller gravitational pull. To help explain his theories of gravity and motion, Newton helped create a new, specialized form of mathematics.

Explanation:Here I dont know

Answer:

Explanation:

Using the equation of motion S = ut + 1/2at² to get the height of drop where

u is the initial velocity of the object = 0m/s

a is the acceleration due to gravity = +9.81m/s² (downwards motion of object)

t is the time it takes the object to drop = 2secs

Substituting the given parameters into the formula to get the height of drop S, we will have;

S = 0(2)+1/2(9.81)(2)²

S = 0+9.81*2

S = 19.62m

Hence the object will drop at a distance of 19.62m

b.) To determine the time it will take a 5 kg object to drop the same distance, we will use the same formula S = ut+1/2at²

Since the 5kg object also drops at the same distance, then S = 19.62m

Substituting this values into the equation we will have;

19.62 = 0(t) + 1/2(9.81)t²

19.62 = 4.905t²

t² = 19.62/4.905

t² = 4

t =2secs

This shows that it will take 5kg object 2secs to fall from the  same distance. This means that no matter the mass of the object, it will take them the same time to fall at the same distance because they are all falling under the same influence of gravity.

Answer:

Yes

Explanation:

This is because the interaction between piece of paper and earth.is gravitational while that of piece of paper and rod is electrostatic

Yes, the interaction between a piece of paper and the rod, stronger than the gravitational interaction between the piece of paper and the Earth.

Therefore, the interaction between a piece of paper and the rod, stronger than the gravitational interaction between the piece of paper and the Earth.

Learn more:

https://brainly.com/question/17907327

Answer:

The force between the charges are not affected.

Explanation:

Given;

distance between two equal charges, R = 2m

The force between the charges is given by;

[tex]F = \frac{kq^2}{R^2}\\\\F_1 = \frac{kq_1^2}{R_1^2}\\\\When\ the \ charges \ and \ the \ distance \ are \ divided \ by \ two \ (q_2 = \frac{q_1}{2}, \ R_2 = \frac{R_1}{2} )\\\\ F_2 = \frac{kq_2^2}{R_2^2}\\\\F_2 = \frac{k(q_1/2)^2}{(R_1/2)^2}\\\\F_2= \frac{4k*q_1^2}{4*R_1^2}\\\\F_2 = \frac{k*q_1^2}{R_1^2}\\\\F_2 = F_1[/tex]

Therefore, the force between the charges are not affected.

Explanation:

binary number is a number expressed in the base-2 numeral system or binary numeral system, which uses only two symbols: typically "0" and "1".

Answer:

a) 121 km

b) 74°

Explanation:

To start with, we assume that there exist two components, the East and the North. We'd be representing the East, by "e" and the North, by "n"

Now we start with the

First vector:

east1 = 232 cos 30 = 201

north1 = 232 sin 30 = 116

Now, that of the second vector will be

east 2 = - 168

north2 = 0

Next, we add the two together and get

East components

201 - 168 = 33 east

North components

116 + 0 = 116

Therefore, the magnitude has to be

magnitude = √(33² + 116²)

Magnitude = √14545

Magnitude = 121

tanθ = 116/33

Tanθ = 3.51

θ = tan^-1 3.51

θ = 74° North East

Question:

What quantity do units represent in a value? A. Size B. Direction C. Magnitude D. Dimension

Answer:

D. Dimension

Explanation:

A unit is simply the measurement given to a dimension. In other words, units represent dimensions. For example, consider a ruler that is 20m long.

i. The value is 20m.

ii. The magnitude of the measurement is 20

iii. The unit of the measurement is meters(m) and that essentially represents the dimension (length) of the ruler.

Answer:

The ball will be overtake superman after travelled distance 248.5 m.

Explanation:

Given that,

Speed of superman = 35 m/s

Height = 330 m

Mass of ball = 10 kg

Let the height at which they are at same position be S.

For superman,

We need to calculate the time

Using equation of motion

[tex]S=ut[/tex]...(I)

For ball,

[tex]S=u't+\dfrac{1}{2}gt^2[/tex]

From equation (I) and (II)

[tex]ut=u't+\dfrac{1}{2}gt^2[/tex]

[tex]ut=0+\dfrac{1}{2}\times g\times t^2[/tex]

[tex]u=\dfrac{1}{2}\times g\times t[/tex]

[tex]t=\dfrac{2u}{g}[/tex]

Put the value into the formula

[tex]t=\dfrac{2\times35}{9.8}[/tex]

[tex]t=7.1\ sec[/tex]

We need to calculate the distance travelled

Using formula of distance

[tex]d=v\times t[/tex]

Put the value into the formula

[tex]d=35\times 7.1[/tex]

[tex]d=248.5\ m[/tex]

Hence, The ball will be overtake superman after travelled distance 248.5 m.

Answer:

Ok, let's suppose the simplest of the physical changes:

We have an object that is not moving (so it is not accelerated)

and there is change, now the object moves.

Because there was a change, means that there was an acceleration, and by the second Newton's law.

Force equals mass times acceleration:

F = m*a

There must be a force.

So suppose that you pushed the object, then some energy that you had, you transferred it to the object, that now is moving and now has kinetic energy.

Now, is kinda true that in a closed system the total energy is always constant, but it depends on what is our system.

So if we think in our system as you and the object, then in the whole system the energy does not change because the energy that you lost is now on the object, but again, there was a transfer of energy.

So no, your friend is not correct.

Both chemical and physical changes involve energy transfer, so your friend is not correct.

A chemical change occurs when a new substance is formed or created through a process of chemical reaction which is reversible.

The addition or removal of heat energy can affect the rate of chemical reaction. This addition or removal is known as energy transfer process.

Examples of chemical changes include;

A physical change on the other hand doesn't involve formation of new substance and it is can be reversible or irreversible.

The addition or removal of heat energy can affect the rate of physical changes.

Examples of such physical changes include;

Thus, both process (chemical and physical changes) involve energy transfer, so your friend is not correct.

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The area of ANY circle is (π) · (radius²).

So ...

(π) · (radius²) = 154 cm²

radius² = (154 cm²) / (π)

radius² = 49.02 cm²

radius = √(49.02 cm²)

radius = 7 cm

Answer:

[tex] \boxed{\sf Radius \ of \ circle \ (r) = 7 \ cm} [/tex]

Given:

Area of circle = 154 cm²

To Find:

Radius of circle (r)

Explanation:

[tex] \bold{Area \: of \: circle = \pi r^2}[/tex]

[tex] \sf \implies \pi {r}^{2} = 154 \\ \\ \sf \implies {r}^{2} = \frac{154}{\pi} \\ \\ \sf \implies {r}^{2} = \frac{154}{ \frac{22}{7} } \\ \\ \sf \implies {r}^{2} = \frac{154 \times 7}{22} \\ \\ \sf \implies {r}^{2} = \frac{1078}{22} \\ \\ \sf \implies {r}^{2} = 49 \\ \\ \sf \implies {r}^{2} = {7}^{2} \\ \\ \sf \implies r = \sqrt{ {7}^{2} } \\ \\ \sf \implies r = 7 \: cm[/tex]

The question is incomplete. Here is the complete question.

The map (in the attachment) shows Olivia's trip to the coffee shop. She gets on her bike at Loomis and then rides south 0.9mi to Broadway. She turns east onto Broadway, rides 0.8mi to where Broadway turns, and then continues another 1.4mi to the shop.

What is the magnitude of the total displacement of her trip?

Whta is the direction of the total displacement of her trip?

Answer: Magnitude = 2.6mi

              Direction: 54.65° east

Explanation: Displacement is the change in postition of a moving object.

There are a few ways to determine total displacement. For this case, the Perpendicular Components of a Vector method will be used.

For this method, total displacement is given by:

[tex]\Delta d_{t}=\sqrt{(\Delta d_{x})^{2}+(\Delta d_{y})^{2}}[/tex]

[tex]\theta=tan^{-1}(\frac{\Delta d_{y}}{\Delta d_{x}})[/tex]

[tex]\Delta d_{x}[/tex] is the x-component of total displacement and it is the sum of each individual x-components;

[tex]\Delta d_{y}[/tex] is the y-component of total displacement and it is the sum of each individual y-components;

θ is the angle the resulting displacement;

For Olivia's trip, there are no x-component of the first part and for the third part, the path she bikes is a hypotenuse of a right triangle. So, that right triangle's x-component is:

[tex]sin30=\frac{x}{1.4}[/tex]

[tex]\frac{1}{2} =\frac{x}{1.4}[/tex]

x = 0.7

Then,

[tex]\Delta d_{x}[/tex] = 0 + 0.8 + 0.7

[tex]\Delta d_{x}[/tex] = 1.5

Related to y, there are no y-component in the second part of Olivia's trip and for the third part:

[tex]cos30=\frac{y}{1.4}[/tex]

[tex]\frac{\sqrt{3} }{2} =\frac{y}{1.4}[/tex]

y = 1.21

Then,

[tex]\Delta d_{y}[/tex] = 0.9 + 0 + 1.21

[tex]\Delta d_{y}[/tex] = 2.11

Total displacement is

[tex]\Delta d_{t}=\sqrt{(1.5)^{2}+(2.11)^{2}}[/tex]

[tex]\Delta d_{t}=\sqrt{6.7021}[/tex]

[tex]\Delta d_{t}=[/tex] 2.6

Magnitude of Olivia's total displacement is 2.6mi

On the map, joining the initial and final points gives a vector pointing towards east at angle:

[tex]\theta=tan^{-1}(\frac{2.11}{1.5})[/tex]

[tex]\theta=tan^{-1}(1.41)[/tex]

θ = 54.65°

Direction of total displacement is 54.65° East.

Answer:

1.9 mi, 350.7°

Explanation:

If continued another 1.2 mi*

solving for x

0 + 0.8 + 1.2(cos30) = 1.83923048 --> 1.84

solving for y

-0.9 + 0 + 1.2(sin30) = -0.3

^negative because it is going downward

a) solving for magnitude

[tex]\sqrt{(1.84)^2+(-0.3)^2} = \sqrt{3.4756} = 1.86429611... = 1.9 mi[/tex]

b) solving for direction of total displacement

[tex]tan^-1 = (\frac{x}{y} ) \\tan^-1 = (\frac{-0.3}{1.84}) = -0.16 \\[/tex]

∘, measured counterclockwise from the eastward direction

360 - 0.16 = 359.84°

*replace any of the needed values in the equation, such as 1.2 mi to 1.4 mi

Answer: Heart rate provides an objective measurement of how hard your body is working. The higher the exercise intensity, the higher your heart rate will be.

Explanation:

Answer:

You're getting exercise and training your heart!

Explanation:

Answer:

0.68 m

Explanation:

Since the volume of water is 310 L, and we know that 1 litre = 1 dm³. So the volume of water is V = 310 dm³. Since this volume of water is the volume of water the cubical steel tank can contain, it equals the volume of the cubical steel tank.

We know that the volume of the cubical steel tank V = L³ where L is the length of side of the cube on the inside.

So the length of side of the cube L = ∛V = ∛310 dm³ = 6.77 dm = 6.77 dm × 1m/10 dm = 0.677 m ≅ 0.68 m

So, the smallest possible inside length of the tank is 0.68 m

Answer: Hydrogen Peroxide

Explanation: Hydrogen peroxide breaks down into oxygen and water. As a small amount of hydrogen peroxide generates a large volume of oxygen, the oxygen quickly pushes out of the container. The soapy water traps the oxygen, creating bubbles, and turns into foam.

Answer:

I think its A.........

Answer:

Hearing, Vision and Metabolism.

(a) For the first 6.2 s, the car has velocity at time t given by

[tex]v(t)=13.5\dfrac{\rm m}{\rm s}+\left(1.9\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]

so that after 6.2 s, it attains a velocity of

[tex]v(6.2\,\mathrm s)=25.28\dfrac{\rm m}{\rm s}[/tex]

For any time t after 6.2 s, its velocity is given by

[tex]v(t)=25.28\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]

which tells us the velocity only falls from this point onward. This means the maximum speed is 25.28 m/s, or about 25.3 m/s.

(b) Solve for t (after 6.2 s) that makes v(t) = 0 :

[tex]25.28\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t=0[/tex]

[tex]\implies t\approx21.067\,\mathrm s[/tex]

It takes the car about 21.2 s to come to a rest, so the car travels a total of about 6.2 s + 21.2 s = 27.4 s.

(c) For the first 6.2 s, the car undergoes a displacement at time t of

[tex]x(t)=\left(13.5\dfrac{\rm m}{\rm s}\right)(6.2\,\mathrm s)+\dfrac12\left(1.9\dfrac{\rm m}{\mathrm s^2}\right)(6.2\,\mathrm s)^2[/tex]

[tex]\implies x\approx120.218\,\mathrm m[/tex]

For time t beyond 6.2 s, its displacement is

[tex]x(t)=120.218\,\mathrm m+\left(25.28\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]

The car comes to a rest after 21.2 s (accelerating at a rate of -1.2 m/s^2), so that its total displacement is

[tex]x(21.2\,\mathrm s)\approx386.49\,\mathrm m[/tex]

so the car travels a total distance of about 387 m.