2 years ago

Placed exactly between two oppositely charged point charges, a test charge (the

Physics

1 answer:

Sveta_85 [38]2 years ago

6 0

Zero maximum force (N) or field strength (N/C). ... minimum /maximum field strength.

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If a ball is launched horizontally at 40 m/s

vaieri [72.5K]

Answer:

40m/s

Explanation:

The horizontal component of velocity remains constant because there are no external forces in that direction

By applying motion equations, V= U+ at

where ,

v - final velocity
u - initial velocity
a-acceleration
t - time

v = u +at

As no force act on the ball ( we neglect air resistance here) no acceleration is seen,

So v = u = 40m/s

4 0

2 years ago

A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis (Fig. 11-48).A toy train

masya89 [10]

Answer:

0.166 rad/s

Explanation:

See attachment for calculations

5 0

2 years ago

Read 2 more answers

Two flywheels of negligible mass and different radii are bonded together and rotate about a common axis (see below). The smaller

jekas [21]

Answer:

$F_2 = 29.54 N$

Explanation:

As we know that the combination is maintained at rest position

So we will take net torque on the system to be ZERO

so we know that

$\tau = \vec r \times \vec F$

here we will have

$\vec r_1 \times F_1 = \vec r_2 \times F_2$

so we have

$13 \times 50 = 22 \times F_2$

so we have

$F_2 = \frac{13 \times 50}{22}$

$F_2 = 29.54 N$

8 0

2 years ago

Blocks A (mass 2.00 kg) and B (mass 10.00 kg, to the right of A) move on a frictionless, horizontal surface. Initially, block B

olga2289 [7]

Answer:dont care

Explanation:

5 0

3 years ago

A physics student stands on the rim of the canyon and drops a rock. The student measures the time for it to reach the bottom to

zysi [14]

Answer:

50.2 m

Explanation:

We can solve the problem by using the following SUVAT equation for the vertical position of the rock:

$y(t)=h+ut+\frac{1}{2}gt^2$

where

h is the initial height (the depth of the canyon), taking the bottom of the canyon as reference position

u = 0 is the initial velocity of the rock

t is the time

$g=-9.8 m/s^2$ is the acceleration of gravity

When the rock reaches the bottom, t = 3.2 s and y = 0. Substituting these numbers and solving for h, we find the depth of the canyon:

$h=\frac{1}{2}gt^2 = -\frac{1}{2}(-9.8)(3.2)^2=50.2 m$

5 0

2 years ago