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2 years ago

Calculate the work done when a 10 N force pushes against a cart 4.0 m

Physics

2 answers:

juin [17]2 years ago

7 0

work = force x distance

10N and 4m gives 40 joules

Pavel [41]2 years ago

4 0

Hello There!

The answer is provided in the image attached.

Have a great day!

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Two 2.0 kg bodies, A and B, collide. The velocities before the collision are ~vA = (15ˆi + 30ˆj) m/s and ~vB = (−10ˆi + 5.0ˆj) m

AleksandrR [38]

Answer:

Part a)

$10\hat i + 15\hat j = \vec v$

Part b)

$\Delta K = 500 J$

Explanation:

As we know that there is no external force on the system of two masses so here total momentum of the system will remains conserved

so we can say

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

$(2kg)(15\hat i + 30 \hat j) + (2 kg)(-10\hat i + 5\hat j) = 2kg(-5\hat i + 20\hat j) + 2\vec v$

$5\hat i + 35\hat j = (-5\hat i + 20\hat j) +\vec v$

$10\hat i + 15\hat j = \vec v$

Part b)

magnitude of the initial speed of A = $\sqrt{15^2 + 30^2} = 33.54 m/s$

magnitude of the initial speed of B = $\sqrt{10^2 + 5^2} = 11.18 m/s$

magnitude of final speed of A = $\sqrt{5^2 + 20^2} = 20.61 m/s$

magnitude of final speed of B = $\sqrt{10^2 + 15^2} = 18.03 m/s$

Now change in total kinetic energy is given as

$\Delta K = (\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2) - (\frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2)$

$\Delta K = (\frac{1}{2}2(33.54)^2 + \frac{1}{2}2(11.18)^2) - (\frac{1}{2}2(20.61)^2 + \frac{1}{2}2(18.03)^2)$

$\Delta K = 500 J$

6 0

3 years ago

shepuryov [24]

I think it's three days. I read it in assignment potion before but it's kinds fuzzy but I believe it's three days. Hopefully thats correct.

4 0

3 years ago

A car is traveling at a velocity of 22 m/s when the driver puts on the brakes

Brums [2.3K]

The car’s velocity at the end of this distance is 18.17 m/s.

Given the following data:

Initial velocity, U = 22 m/s

Deceleration, d = 1.4 $m/s^2$

Distance, S = 110 meters

To find the car’s velocity at the end of this distance, we would use the third equation of motion;

Mathematically, the third equation of motion is calculated by using the formula;

$V^2 = U^2 + 2dS$

Substituting the values into the formula, we have;

$V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}$

Final velocity, V = 18.17 m/s

Therefore, the car’s velocity at the end of this distance is 18.17 m/s.

Read more: brainly.com/question/8898885

8 0

1 year ago

A 30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force \vec{F} F ⃗ so that the swing ropes

Virty [35]

Answer:

169.74 N

Explanation:

Given,

Mass of the girl = 30 Kg

angle of the rope with vertical, θ = 30°

equating the vertical component of the tension

vertical component of the tension is equal to the weight of the girl.

T cos θ = m g

T cos 30° = 30 x 9.8

T = 339.48 N

Tension on the two ropes is equal to 339.48 N

Tension in each of the rope = T/2

= 339.48/2 = 169.74 N

Hence, the tension in each of the rope is equal to 169.74 N

7 0

2 years ago

Starting with the definition 1.00 in. = 2.54 cm, find the number of kilometers in 8.00 mi .

velikii [3]

Use Factor-Label Method:

8miles 63360 inches
---------- X --------------------- X
1 1 mile

2.54cm 1 meter
X ------------ X ---------------- X
1 inch 100 cm

1 km
----------------- = 12.87 km
1000meters

8 miles = 12.87 km

8 0

3 years ago