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ANEK [815]
4 years ago
9

Here is the situation: A puck is resting on the floor of a large moving van. Assume that the floor of the van is frictionless. T

he van and the puck are initially at rest, then the van undergoes a positive uniform acceleration. Determine whether each statement below is:a. True b. False.
Physics
1 answer:
Ahat [919]4 years ago
3 0

Answer:

Statement 1: Someone outside the van sees the puck remain stationary with respect to the ground.

TRUE. This is because the puck relative to the ground initially was stationary. When the van starts to move forward or accelerate, the puck will tend to maintain its stationary state with the ground by moving in the opposite direction. This movement is only relative to the van and not the ground because the puck and the ground are at rest as the puck has no net force acting on it.

Someone in the van sees the puck move toward the front of the van.

FALSE. The puck has no net force acting in it so by newton's first law it should have no relative movement to the ground. In order to achieve this, the puck by newton's third law of motion must move in the opposite direction (backwards) as the van to maintain its stationary state with the ground.

The acceleration of the van with respect to the ground is > the acceleration of the puck with respect to the ground.

If the van turns to the right, the puck will hit the left wall.

WHAT DOES THIS HAVE TO DO WITH THE STATEMENTS? I'll assume the truck continues in a straight line and this statement if irrelevant. It does say "If"

The magnitude of the acceleration of the van with respect to the ground is = the magnitude of the puck's acceleration with respect to the van.

The puck is stationary with respect to the van during this acceleration.

FALSE

Explanation:

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an aircraft is flying at height of 3400m above the ground.if the angle subtended at a ground observation point by the aircraft p
Kamila [148]

Answer: 588.9 m/s

Explanation:

Given that :

θ = 30°

Height, h = 3400m

Time, t = 10 seconds

From trigonometry ;

Tanθ = opposite / hypotenus

Tan 30 = 3400 / x

x tan 30 = 3400

0.5773502x = 3400

x = 3400 / 0.5773502

x = 5888.9727

Recall ;

Speed = Distance / time

Speed = 5888.9727 / 10

Speed = 588.897 m/s

Speed = 588.9 m/s

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3 years ago
Q Explain the following terms.<br> a) A source of electric current<br> b) Electrical appliances
exis [7]

Answer:a)battery

B)ironer

Explanation:

4 0
2 years ago
The mass of the Mars is about 0.11 the mass of the earth, its radius is 0.53 that of the earth, and the acceleration due to grav
pishuonlain [190]

Answer:

3.8377m/s^2

Explanation:

The force due to gravity on earth's surface on mass m is:

F=GMm/R^2

Where,

F = GMm/R2

F = Gravitational Force

G = Gravitational Constant

M = Earth's Mass

m = mass of body

R = Earth's radius

The acceleration is F/m:

g=F/m=GM/R^2

For a planet with 0.11M and 0.53R:

a=G(0.11M)/(0.53R)^2\\=0.3915g\ \ \ \ \ \ \g=9.8m/s^2\\=0.3915\times9.8m/s^2\\\\=3.8377m/s^2

Hence gravity on Mar's surface is 3.8377m/s^2

4 0
3 years ago
16. How much force would it take to accelerate a 75kg object by 5 m/szą
patriot [66]

Answer:

The force experienced is 375 N.

Explanation:

Here,

Mass of the object( M)= 75 kg

Acceleration of the body(a) =5 m/s²

Force required to accelerate the body( F) =?

Then, we know ,

From newton's second law of motion,

F=M*a

     =75* 5

    =375 kgm/s²

   = 375 Newton

So, The force experienced by the 75 kg body that undergoes an acceleration of 5 m/s² is 375 N.

7 0
3 years ago
A physics major is working to pay his college tuition by performing in a traveling carnival. He rides a motorcycle inside a holl
Semenov [28]

Answer:

a) v=23.9 m/s

b) R=2646 N

Explanation:

According to Newton's second law the net force at the top is given by

∑F=-m*ac

according to

a_{c}=\frac{v^2}{r}

∑F=-m*\frac{v^2}{r}

a).

To lose contact this means that R=0 so the final equation is

R-m*g=-m*\frac{v^2}{r}

-m*g=-m*\frac{v^2}{r}

Solve to v

v^2=g*r

v=\sqrt{g*r}=\sqrt{9.8*14.6}

v=11.96\frac{m}{s}

b).

v is the twice of part a so

v=2*11.96

v=23.9\frac{m}{s}

R_{m}-m_{m}*g-m_{p}=m*\frac{v^2}{r}

Solve to Rm

R_{m}=m*\frac{v^2}{r}+(m_{m}+m_{p})*g

R_{m}=m*\frac{23.9^2}{14.6}+(40+70)*9.8

R_{m}=2646 N

5 0
3 years ago
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