PLZ HELP!! WILL MARK BRAINLIEST!!

A 75.0-kg painter climbs a ladder that is 2.75 m long leaning against a vertical wall. The ladder makes a 30.0° angle with the w

Schach [20]

<h3><u>Answer;</u></h3>

a) 178.6125 Joules

b) Work done does not depend on constant speed or acceleration up the ladder

<h3><u>Explanation</u>;</h3>

Work done =Fd times the cosine of angle between;

where F is force d is distance m is the mass a is acceleration

.F=ma. In this case a=g=9.81 m/s^2 and cos 30=0.866

Therefore;

W= 75 × 9.81 × 2.75 × .866

= 178.6125 Joules

b. The work done by the painter does not depend on whether the painter climbs at constant speed or accelerates up the ladder.

W=3376.945 Joules

6 0

3 years ago

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Problem 7:__deletededc326999bef85931fda5d1ab0d68e21218f27b46f4f11ab80872358251b584ddeleted__ 0N86-C1-52-40-A837-22820 If object

laila [671]

Answer:

the kinetic energy of body B is twice the kinetic energy of body A

Explanation:

The kinetic energy of a body is given by

K = ½ m v²

If two objects leave the same point, suppose that at the same height when they reach the ground they have the same velocity.

Therefore if the mass of body b is twice the mass of body A

$m_{b} = 2 m_{a}$

$K_{b}$ = ½ (2 $m_{a}$ ) v²

K_{b} = 2 (½ m_{a} v²)

K_{b} = 2 K_{a}

therefore the kinetic energy of body B is twice the kinetic energy of body A

3 0

3 years ago

What is a zippper<br> awsedrftvgbhnjmk,l

lions [1.4K]

Answer:

a jacket closing mechanism a most useful tool to preserve and protect.

Explanation:

4 0

3 years ago

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3) A driver in a 1000-kg car traveling at 24 m/s slams on the brakes and skids to a stop. If the coefticient of friction between

Natali5045456 [20]

Answer:

A) 37 m

Explanation:

The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:

$v^2 -u^2 = 2ad$ (1)

where

v = 0 m/s is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d is the length of the skid

We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:

$F=ma=-\mu mg$

where

m = 1000 kg is the mass of the car

$\mu = 0.80$ is the coefficient of friction

a is the deceleration of the car

g = 9.8 m/s^2 is the acceleration due to gravity

The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:

$a=-\mu g$

And we can substitute it into eq.(1) to find d:

$v^2 -u^2 = 2(\-mu g) d\\d= \frac{v^2-u^2}{-2 \mu g}=\frac{0-(24 m/s)^2}{-2(0.80)(9.8 m/s^2)}=36.7 m \sim 37 m$

7 0

3 years ago

Problem 2: (15 pts) A 10-m high cylindrical container is half-filled at the bottom with water of density =1000 kg/m3 while the t

Step2247 [10]

Answer:

$\Delta p = 90.7 kPa$

Explanation:

specific gravity of oil is $= \frac{\rho_{oil}}{\rho_w}$

$\rho_{oil} = 0.85*1000 = 850 kg/m3$

we know that

change in pressure for oil is given as

$\Delta p = \rho gh$

here density and h is for oil

$\Delta p = 850*5 *9.81 = 41,692.5 kPa$

change in pressure for WATER is given as

$\Delta p = \rho gh$

here density is for water and h is for water

$\Delta p = 1000*5 *9.81 = 49,050 kPa$

pressure change due to both is given as

$\Delta p = 41692.3 + 49050 = 90742.5 N/m2$

$\Delta p = 90.7 kPa$

8 0

3 years ago

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