What is the momentum of a 56-kilogram ice skater gliding across the ice at a speed of 12 m/s? kg-m/s

When a point charge of q is placed on one corner of a square, an electric field strength of 2 N/C is observed at the center of t

bearhunter [10]

Answer:

Explanation:

Net electric field at the centre will be zero .

Since all the charges are equal and they all are symmetrically situated around the centre . So the electric field produced by each will cancel out each other and hence the resultant electric field will be zero . It happens because electric field is a vector quantity and therefore it adds up vectorially . All the four electric field will form two pairs , in each pair electric fields are acting in opposite direction . So they all cancel out to zero .

3 0

3 years ago

A child of mass 25 kg is skating fast, +10 m/s, and tries to get revenge by colliding with the 60 kg adult who is sitting still.

tankabanditka [31]

Answer: -4.4 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1}V_{o}+m_{2}U_{o}$ (2)

$p_{f}=m_{1}V_{f}+m_{2}U_{f}$ (3)

$m_{1}=25 kg$ is the mass of the child

$V_{o}=10 m/s$ is the initial velocity of the child

$m_{2}=60 kg$ is the mass of the adult

$U_{o}=0 m/s$ is the initial velocity of the adult (it is sitting still)

$V_{f}$ is the final velocity of the child

$U_{f}=6 m/s$ is the final velocity of the adult

Substituting (2) and (3) in (1):

$m_{1}V_{o}+m_{2}U_{o}=m_{1}V_{f}+m_{2}U_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1}V_{o}-m_{2}U_{f}}{m_{1}}$ (5)

$V_{f}=\frac{(25 kg)(10 m/s)-(60 kg)(6 m/s)}{25 kg}$ (6)

Finally:

$V_{f}=-4.4 m/s$ This means the velocity of the child is in the opposite direction

4 0

3 years ago

Lf you divide a speed in miles per minute by 60, you get the same speed<br> expressed in miles per

Zielflug [23.3K]

Ok that is interesting.

8 0

3 years ago

A wooden bucket filled with water has a mass of 68 kg and is attached to a rope that is wound around a cylinder with a radius of

Aleonysh [2.5K]

Answer: 210.2N

Explanation:

Assume a bucket of water with a total mass of 68kg is attached to a rope, which in turn is tied around a 0.078m radius cylinder at the top of a well. A crank with a turning radius of 0.250 m is attached to the end of the cylinder.

the minimum force directed perpendicular to the crank handle required to raise the bucket is

(Assume the rope's mass is negligible, that the cylinder turns on friction-less bearings, and that g = 9.8 m/s2

The crank handle provides a torque T=0.25F where F is the force we are looking for.

A free body diagram will show that the tension in the rope times the cylinder radius R is equal to the torque on the cylinder. But the tension in the rope is just the weight of the bucket

W=mg= 68kg

W(0.078)=T=0.25F

F=0.312W=0.312(68kg)=21.216kg= 210.2N

7 0

3 years ago

A 0.145-kg baseball pitched horizontally at 35.0 m/s strikes a bat and is popped straight up to a height of 40.0m. If the contac

Nookie1986 [14]

Answer:

$F = 2.6 \times 10^3 N$