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4 years ago

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Physics

2 answers:

alexgriva [62]4 years ago

4 0

Answer:

w and z

Explanation:

multiply the forces and distances together to solve for work

first and last ones give you 1125

Got it right on Edgenuity

Novosadov [1.4K]4 years ago

3 0

Answer:

W and Z

Explanation:

Work done is the force applied to move a body through a distance in a specific direction.

Work done = Force x distance

Now let's use this formula to solve for work done;

Scenario W: A force of 75 N moves a box 15 m.

Work done = 75 x 15 = 1125J

Scenario X: A force of 100 N moves a box 12 m.

Work done = 100 x 12 = 1200J

Scenario Y: A force of 50 N moves a box 20 m.

Work done = 50 x 20 = 1000J

Scenario Z: A force of 25 N moves a box 45 m.

Work done = 25 x 45 = 1125J

Choices W and Z is the right option in which the same amount of work is done on the box.

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Which trial’s cart has the greatest momentum at the bottom of the ramp?

Licemer1 [7]

The momentum of each cart is given by:
$p=mv$
where
m is the mass of the cart
v is its velocity (at the bottom of the ramp)

To answer the problem, let's calculate the momentum of each of the 4 carts:
1) $p=(200 kg)(6.5 m/s)=1300 kg m/s$
2) $p=(220 kg)(5.0 m/s)=1100 kg m/s$
3) $p=(240 kg)(6.4 m/s)= 1536 kg m/s$
4) $p=(260 kg)(4.8 m/s)=1248 kg m/s$

Therefore, the cart with greatest momentum is cart 3, so the right answer is
<span>- trial 3, because this trial has a large mass and a large velocity</span>

8 0

3 years ago

Sarah, who has a mass of 55 kg, is riding in a car at 20 m/s. She sees a cat crossing the street and slams on the brakes! Her se

lions [1.4K]

Answer:

-2200 N

Explanation:

Here we can use the impulse theorem, which states that the impulse exerted on Sarah (product of force and duration of collision) is equal to Sarah's change in momentum:

$I=\Delta p\\F \Delta t = m \Delta v$

where

F is the average force

$\Delta t$ is the duration of the collision

m is the mass

$\Delta v$ is the change in velocity

In this problem:

m = 55 kg

$\Delta v = 0-20 = -20 m/s$

$\Delta t = 0.5 s$

Solving the formula, we find the force exerted by the seatbelt on Sarah:

$F=\frac{m\Delta v}{\Delta t}=\frac{(55)(-20)}{0.5}=-2200 N$

And the negative sign means the direction is opposite to that of Sarah's initial motion.

6 0

3 years ago

Scientists observe an approaching asteroid that is on a collision course with

nasty-shy [4]

Answer:

The approximate velocity the rocket must have to stop the asteroid completely after the collision is;

C. -324 m/s

Explanation:

The parameters of the asteroid and the rocket are;

The mass of the asteroid, m₁ = 11,000 kg

The initial velocity with which the asteroid is approaching Earth, v₁ = 50 m/s

The mass of the rocket, m₂ = 1700 kg

The initial velocity of the rocket = v₂

The final velocity of the combined asteroid and rocket after the collision, v₃ = 0 m/s

By the law of conservation of linear momentum, we have;

The total initial momentum = The total final momentum

m₁·v₁ + m₂·v₂ = (m₁ + m₂)·v₃

Substituting the known values, we get;

11,000 kg × 50 m/s + 1,700 kg × v₂ = (11,000 kg + 1,700 kg) × 0 m/s

11,000 kg × 50 m/s + 1,700 kg × v₂ = 0

∴ 1,700 kg × v₂ = -11,000 kg × 50 m/s

v₂ = (-11,000 kg × 50 m/s)/(1,700 kg) = -323.529412 m/s ≈ -324 m/s

The approximate initial velocity the jet must have to completely stop the asteroid after the collision is -324 m/s.

3 0

3 years ago

Given vectors D (3.00 m, 315 degrees wrt x-axis) and E (4.50 m, 53.0 degrees wrt x-axis), find the resultant R= D + E. (a) Write

Eva8 [605]

Answer:

R = ( 4.831 m , 1.469 m )

Magnitude of R = 5.049 m

Direction of R relative to the x axis= 16°54'33'

Explanation:

Knowing the magnitude and directions relative to the x axis, we can find the Cartesian representation of the vectors using the formula

$\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ its the magnitude and θ.

So, for our vectors, we will have:

$\vec{D}= 3.00 m \ ( \ cos(315) \ , \ sin (315) \ )$

$\vec{D}= ( 2.121 m , -2.121 m )$

and

$\vec{E}= 4.50 m \ ( \ cos(53.0) \ , \ sin (53.0) \ )$

$\vec{E}= ( 2.71 m , 3.59 m )$

Now, we can take the sum of the vectors

$\vec{R} = \vec{D} + \vec{E}$

$\vec{R} = ( 2.121 \ m , -2.121 \ m ) + ( 2.71 \ m , 3.59 \ m )$

$\vec{R} = ( 2.121 \ m + 2.71 \ m , -2.121 \ m + 3.59 \ m )$

$\vec{R} = ( 4.831 \ m , 1.469 \ m )$

This is R in Cartesian representation, now, to find the magnitude we can use the Pythagorean theorem

$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$

$|\vec{R}| = \sqrt{(4.831 m)^2 + (1.469 m)^2}$

$|\vec{R}| = \sqrt{23.338 m^2 + 2.158 m^2}$

$|\vec{R}| = \sqrt{25.496 m^2}$

$|\vec{R}| = 5.049 m$

To find the direction, we can use

$\theta = arctan(\frac{R_y}{R_x})$

$\theta = arctan(\frac{1.469 \ m}{4.831 \ m})$

$\theta = arctan(0.304)$

$\theta = 16\°54'33''$

As we are in the first quadrant, this is relative to the x axis.

3 0

4 years ago

All objects within a closed system tend to move

Charra [1.4K]

Explanation is in the file

tinyurl.com/wpazsebu

4 0

3 years ago