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Doss [256]
4 years ago
13

You push a heavy crate out of a carpeted room and down a hallway with a waxed linoleum floor. While pushing the crate 2.5 m out

of the room you exert a force of 385 N; while pushing it 10 m down the hallway you exert a force of 130 N. To slow it down you exert a force of 350 N through a distance of 2.0 m, opposite to the motion. How much work do you do in all?
Physics
1 answer:
Alinara [238K]4 years ago
3 0

Answer:

W = 1562.5 J

Explanation:

Path 1:

W₁ = F₁*d₁ = 385 N * 2.5 m = 962.5 J

Path 2:

W₂ = F₂*d₂ = 130 N * 10 m = 1300 J

Path 3:

W₃ = F₃*d₃ = (-350 N) * 2 m = - 700 J  (opposite to the motion)

We get

W = W₁ + W₂ + W₃ = 962.5 J + 1300 J + (- 700 J) = 1562.5 J

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Answer:

Time, I believe. Pretty sure it's time lol

3 0
3 years ago
How is the motion of an object affected when a force acts on it?
Arada [10]

Newton's first law of motion is sometimes called the law of inertia. When the forces acting on an object are balanced, the object is either at rest or moving with a constant velocity. Unbalanced forces can cause an object to accelerate or decelerate. Unbalanced forces can also cause an object to change direction.

Hope that helps!

3 0
3 years ago
Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the heig
HACTEHA [7]

Answer:

45.6m

Explanation:

The equation for the position y of an object in free fall is:

y=-\frac{1}{2} gt^2+v_0t+y_0

With the given values in the question the equation has one unknown v₀:

v_0=\frac{y-y_0}{t}+\frac{1}{2}gt

Solving for t=1:

1) v_0=y-y_0+\frac{g}{2}

To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

2) E=\frac{1}{2}m{v_0}^2+mgy_0

If h is the height of the tower, the energy on top of the tower:

3) E=mgh

Combining equation 2 and 3 and solving for h:

4) h=\frac{{v_0}^2}{2g}+y_0

Combining equation 1 and 4:

h=\frac{{(y-y_0+\frac{g}{2}})^2}{2g}+y_0

4 0
3 years ago
What is the heat extracted from the cold reservoir for the refrigerator?
zaharov [31]
What is the heat extracted from the cold reservoir for the refrigerator shown in(Figure 1) ? Assume that W1 = -123J and W2 = 88J . 

<span>Qc= _________ </span>

<span>Part B 
</span>
K=105J
7 0
3 years ago
Independent practice
coldgirl [10]

Explanation:

formula: <u>Mass</u>

Density x volume

2a) m=10kg v=0.3m³

10÷0.3=33.3 kg/m

2b) m = 160 kg V=0.1m³

160÷0.1=1600 kg/m

2c) m = 220 kg V = 0.02m³

220÷0.02=11000 kg/m

A wooden post has a volume of 0.025m³ and a mass of 20kg. Calculate its density in kg/m.

density = volume ÷ mass

20÷ 0.025=800 kg/m

Challenge: A rectangular concrete slab is 0.80m long, 0.60 m wide and 0.04m thick. Calculate its volume in m³.

Formula : Length x width x height = Volume

0.80 x 0.60 x 0.04 = 0.0192m³

B) The mass of the concrete slab is 180 kg. Calculate its density in kg/m.

density = volume ÷ mass

180 ÷ 0.0192 = 9375 kg/m

4 0
3 years ago
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