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Len [333]
2 years ago
5

When Jane drives to work, she always places her purse on the passenger’s seat. By the time she gets to work, her purse has falle

n on the floor in front of the passenger seat. One day, she asks you to explain why this happens in terms of physics. What do you say?
Physics
2 answers:
zheka24 [161]2 years ago
6 0

Explanation: There may be different explanations:

Suppose that Jane's car has a built some sped, and she sees a stoplight. Now she needs to stop the car (decelerating the car)

When she does this, she creates a force that holds the car until it is full stop, but this force is applied only on the car, the purse that is on the passenger seat only feels this force by the friction that the seat does in his surface, but the purse has built some moment because it was inside the car, and the friction force that the seat does may not be enough to also stop the purse. So when Jane stops, the purse keeps moving forward, and then it falls down of the seat.

LUCKY_DIMON [66]2 years ago
4 0
At some time during her drive she backed up with a substantial negative. ( backwards) acceleration. Since the pocket book is not physically connected to the seat it is free to move. Upon rapid negative acceleration the pocket book remains in its position while the car accelerates backwards away from it. this demonstrates Newtons 1st law of motion. The first law is the law of inertia. Which states, an object at rest. ( pocketbook) will remain at rest and an object in motion will continue in motion at constant velocity, unless acted upon by some outside force to change its motion.
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Two astronauts, A and B, both with mass of 60Kg, are moving along a straight line in the same direction in a weightless spaceshi
sveta [45]

Answer:

V=14m/s

Explanation:

From the Question we are told that

Mass of A and B is 60kg

Speed of A=2m/s

Speed of B=1m/s

Mass of bag =5kg

Generally the momentum of the astronaut  A and bag is mathematically given as

  M_A=(60+5)*2

   M_A=130kgm/s

Generally to avoid collision the speed of astronaut be should be less than or equal to that of astronaut A with the bag

Therefore for minimum requirement speed of astronaut A should be given by astronaut B's speed which is equal to 1

Therefore

  130=(60*1)=(5*v)

   V=14m/s

7 0
1 year ago
A 5.0 kg cannonball is dropped from the top of a tower. It falls for 1.6 seconds before slamming into a sand pile at the base of
stepan [7]

Answer:

15.7 m/s

Explanation:

The motion of the cannonball is a accelerated motion with constant acceleration g = 9.8 m/s^2 towards the ground (gravitational acceleration). Therefore, the velocity of the ball at time t is given by:

v(t)=u + gt

where

u = 0 is the initial velocity

g = 9.8 m/s^2 is the acceleration

t is the time

If we substitute t=1.6 s into the equation, we find the final velocity of the cannonball:

v(1.6 s)=0+(9.8 m/s^2)(1.6 s)=15.7 m/s

4 0
2 years ago
I pull the throttle in my racing plane at a = 12.0 m/s2. I was originally flying at v = 100. m/s. Where am I when t = 2.0s, t =
Helen [10]
Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)

• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)

•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)
7 0
2 years ago
A child of mass m is at the edge of a merry-go-round of diameter d. When the merry-go-round is rotating with angular acceleratio
dem82 [27]

Answer:

The torque on the child is now the same, τ.

Explanation:

  • It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.
  • In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.
  • The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of  the mass of the child times the square of the distance to the center.
  • When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:

       I_{to} = I_{d} + m*r^{2}  = m*\frac{r^{2}}{2} +  m*r^{2} = \frac{3}{2}*  m*r^{2} (1)

  • So, τ = 3/2*m*r²*α (2)
  • When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:

       I_{t} = I_{d} + m*\frac{r^{2}}{4}   = m*\frac{r^{2}}{2} + m*\frac{r^{2}}{4}  = \frac{3}{4}*  m*r^{2} (3)

  • Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:

       τ = 3/4*m*r² * (2α) = 3/2*m*r²

        same result than in (2), so the torque remains the same.

7 0
2 years ago
What are the reasons for the establishment of UGA?
3241004551 [841]

Answer:

In February 1784, just after the close of the Revolutionary War, the General Assembly of Georgia earmarked 40,000 acres of land to endow "a college or seminary of learning." The following year, Abraham Baldwin, a lawyer and minister educated at Yale University in New Haven, Connecticut, who had settled in Georgia

Explanation:

please mark this answer as brainliest

8 0
1 year ago
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