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Leviafan [203]
3 years ago
5

Gravity does not actually "pull" objects at

Physics
1 answer:
Radda [10]3 years ago
5 0
I believe it’s (D. Any object)
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A 7382 N piano is to be pushed up a(n) 2.16 m frictionless plank that makes an angle of 22.7 ◦ with the horizontal. Calculate th
lubasha [3.4K]

Answer:

Work done, W = 6153.31 Joules

Explanation:

It is given that,

Weight of piano, W = F = 7382 N

It is pushed 2.16 meters friction less plank

The angle with horizontal, \theta=22.7^{\circ}

When the piano slide up plank at a slow constant rate. The y component of force is taken into consideration. The net force acting on it is given by :

F_y=F\ sin\theta

Work done is given by :

W=F_y\times d

W=F\ sin\theta \times d

W=7382 sin(22.7) \times 2.16

W = 6153.31 Joules

So, the work done in sliding the piano up the plank is 6153.31 Joules. Hence, this is the required solution.

7 0
3 years ago
Your neighbor Paul has rented a truck with a loading ramp. The ramp is tilted upward at 25°, and Paul is pulling a large crate u
Alisiya [41]

Explanation:

It is given that,

The ramp is tilted upwards at 25 degrees and Paul is pulling a large crate up the ramp with a rope that angles 10° above the ramp.

Total angle with respect to ramp is 35 degrees.

If Paul pulls with a force of 550 N.

The horizontal component of the force is given by :

F_x=F\ cos\theta

F_x=550\ cos(35)

F_x=450.53\ N

The vertical component of the force is given by :

F_y=F\ sin\theta

F_y=550\ sin(35)

F_y=315.46\ N

Hence, this is the required solution.

4 0
3 years ago
N 1800kg car has an<br> of 3.8m/s? What is it<br> on the car?<br> acceleration<br> force acting
nevsk [136]

Answer:

6840 N

Explanation:

The force acting on the car can be found by using Newton's second law:

F = ma

where

F is the net force on the car

m is the mass of the car

a is its acceleration

For the car in this problem,

m = 1800 kg

a=3.8 m/s^2

Substituting,

F=(1800)(3.8)=6840 N

7 0
3 years ago
Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho
77julia77 [94]

Answer:

The time constant is 1.049.

Explanation:

Given that,

Charge q{t}= 0.65 q_{0}

We need to calculate the time constant

Using expression for charging in a RC circuit

q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]

Where, \dfrac{t}{RC} = time constant

Put the value into the formula

0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]

1-e^{-(\dfrac{t}{RC})}=0.65

e^{-(\dfrac{t}{RC})}=0.35

-\dfrac{t}{RC}=ln (0.35)

-\dfrac{t}{RC}=-1.049

\dfrac{t}{RC}=1.049

Hence, The time constant is 1.049.

6 0
3 years ago
Can any kind soul help me please​
Alex

Answer:

I) 420000J

ii)

Explanation:

(I) so you can use the formula for quantity of heat then substitute the values given

formula-Q=mc∆9

3 0
3 years ago
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