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Arisa [49]
3 years ago
8

How would taping a coin to the balloon affect the overall motion of the balloon?

Physics
1 answer:
Ede4ka [16]3 years ago
3 0

Answer:

the motion of the coin taping the balloon is the balloon squshing down

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We push a 38.4-kg box across the floor at constant velocity. If we are pushing a horizontal force of 238 N, find the coefficient
zmey [24]

Answer: 0.62

Explanation:

Coefficient of friction is defined as the ratio of the moving force (Fm) acting on a body to the normal reaction (R).

Note that the normal reaction acts vertically on the object and is equal to the objects weight (W) i.e W=R

Since W = mg, W = 38.4 ×10

W= 384N =R

Normal reaction = 384N

The horizontal force acting on the body will be the moving force which is 238N

Coefficient of friction = Fm/R

Coefficient of friction = 238/384

Coefficient of friction = 0.62

Therefore, coefficient of kinetic friction between the box and the floor is 0.62

5 0
3 years ago
What would be the weight of the moon if it were resting on the surface of the earth
kari74 [83]
We need to be careful here.
The calculation of the gravitational force between two objects
refers to the distance between their centers. 
The minimum possible distance between the Earth's and moon's
centers is the sum of their radii (radiuses).

Earth's radius . . . . .  6,360 km  =  6.36 x 10⁶ meters
Moon's radius . . . . .  1,738 km  =  1.738 x 10⁶ meters
Sum of their radii  =                      8.098 x 10⁶ meters

Also:
Earth's mass . . . . .  5.972 x 10²⁴ kg
Moon's mass . . . . .  7.348 x 10²²  kg
<span>
and now we're ready to go !

       Gravitational force = 

                   G  M₁ M₂ / R²

= (6.67 x 10⁻¹¹ N-m²/kg²)(</span><span>5.972 x 10²⁴ kg)(7.348 x 10²²  kg)/</span>(8.098 x 10⁶ m)²

= (6.67 · 5.972 · 7.348 / 8.098²) · (10²³)      Newtons

=    (I get ...)        4.463 x 10²³ Newtons

That's almost exactly   10²³ pounds 

                           =  50,153,000,000,000,000,000 tons.     

Those are big numbers. 
All I can say is:  I wouldn't exactly call that "resting" on the surface".
7 0
3 years ago
In a ballistics test, a 24 g bullet traveling horizontally at 1200 m/s goes through a 31-cm-thick 320 kg stationary target and e
Zanzabum

Answer:

The  velocity is  v_t  =  0.02175 \  m/s

Explanation:

From the question we are told that

   The  mass of the bullet is  m_b  =  0.024 \  kg

    The initial speed of the bullet is  u_b  =  1200 \  m/s

   The mass of the target is  m_t  =  320 \  kg

    The  initial velocity of target is  u_t  =  0  \ m/s

    The  final velocity of the bullet is  is  v_b  =  910 \  m/s

   

Generally according to the law of momentum conservation we have that

      m_b *  u_b  +  m_t *  u_t  =  m_b *  v_b  +  m_t  *  v_t

=>   0.024  *  1200  +  320 *  0  =  0.024 *  910   +  320  *  v_t

=>    v_t  =  0.02175 \  m/s

3 0
3 years ago
A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou
Hoochie [10]

Answer:

e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

Explanation:

This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.

The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.

<u>Mathematical expression for the Newton's second law of motion is given as:</u>

F=\frac{dp}{dt} ............................................(1)

where:

dp = change in momentum

dt = time taken to change the momentum

We know, momentum:

p=m.v

Now, equation (1) becomes:

F=\frac{d(m.v)}{dt}

<em>∵mass is constant at speeds v << c (speed of light)</em>

\therefore F=m.\frac{dv}{dt}

and, \frac{dv}{dt} =a

where: a = acceleration

\Rightarrow F=m.a

also

F\propto \frac{1}{dt}

so, more the time, lesser the force.

<em>& </em><u><em>Impulse:</em></u>

I=F.dt

I=m.a.dt

I=m.\frac{dv}{dt}.dt

I=m.dv=dp

∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.

So, the impulse in both the cases will be same.

4 0
3 years ago
For two traveling waves, if the crest of one wave coincides with a trough of another, what type of interference occurs?
kogti [31]
When crest of one wave interferes with the trough of other wave, the amplitude of the resultant wave formed is less. Hence the type of interference is destructive interference.
3 0
3 years ago
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