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4 years ago

13

If the weight of the bowling ball acts down with a force of 200 N, what force would the table need to push up with to keep the b

owling ball from flying skyward or sinking into the table?

1 answer:

zavuch27 [327]4 years ago

3 0

5858585 8 8 855858 858 585858

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What is the charge on an object that experiences a force of 5 Newtons in an electric field of 50 Newtons per coulomb?

andrew-mc [135]

Answer:

Explanation:

F = qE

F is the force in Newtons

q is the test charge

E is the electrical field produced by the source charge

$5=q(50)\\q=1*10^-^1Coulombs$

6 0

3 years ago

Resistors examples.

RUDIKE [14]

Answer:

Multiply the resistances. Example: Two resistors in parallel, 100 and 200 ohms. 100 x 200 = 20,000

Add the resistances. Example: 100 + 200 = 300

Divide the result in Step 1 by the result in Step 2. This gives you the total resistance. Example: 20,000 / 300 = 66.7 ohms.

Explanation:

Multiply the resistances. Example: Two resistors in parallel, 100 and 200 ohms. 100 x 200 = 20,000

Add the resistances. Example: 100 + 200 = 300

Divide the result in Step 1 by the result in Step 2. This gives you the total resistance. Example: 20,000 / 300 = 66.7 ohms.

6 0

3 years ago

A wave with a greater amplitude will transfer . . . . \

Orlov [11]

Answer:

More energy

Explanation:

The amount of energy carried by a wave is related to the amplitude of the wave itself. In particular, the amount of energy carried by the wave is proportional to the square of the amplitude of the wave:

$E \propto A^2$

where

E is the energy

A is the amplitude

This means, for instance, that if the amplitude of a wave is doubled, the energy it carries increases by a factor 4.

8 0

3 years ago

In serving, a tennis player accelerates a 59 g tennis ball horizontally from rest to a speed of 34 m/s Assuming that the acceler

Akimi4 [234]

Answer:

The magnitude of the force exerted on the ball by the racquet is 94.73 N.

Explanation:

The force exerted on the ball is the following:

$F = ma$

Where:

m: is the mass of the ball = 59 g

a: is the acceleration

The acceleration of the ball can be found with the following kinematic equation:

$v_{f}^{2} = v_{0}^{2} + 2ad$

Where:

d: is the distance = 0.36 m

$v_{f}$ : is the final speed = 34 m/s

$v_{0}$ : is the initial speed = 0 (it start from rest)

Hence, the acceleration is:

$a = \frac{v_{f}^{2}}{2d} = \frac{(34 m/s)^{2}}{2*0.36 m} = 1605.6 m/s^{2$

Finally, the force is:

$F = ma = 59 \cdot 10^{-3} kg*1605.6 m/s^{2} = 94.73 N$

Therefore, the magnitude of the force exerted on the ball by the racquet is 94.73 N.

I hope it helps you!

6 0

3 years ago

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x''

Lady_Fox [76]

Answer:

We know that the acceleration of the particle is defined as

$a(t)=\frac{dv}{dt}$

Since it is given that

$v(t)=\frac{5}{\sqrt{t}}\\\\\therefore a(t)=\frac{d(\frac{5}{\sqrt{t}})}{dt}\\\\a(t)=5\times \frac{dt^{-\frac{1}{2}}}{dt}\\\\=\frac{-5}{2}t^{\frac{-1}{2}-1}\\\\\therefore a(t)=x''(t)=\frac{-2.5}{t^{\frac{3}{2}}}$

Now by definition of velocity we have

$v(t)=\frac{dx(t)}{dt}\\\\\Rightarrow dx(t)=v(t)dt$

Integrating on both sides we get

$v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int v(t)dt$

Applying values we get

$v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int \frac{5}{t^{\frac{1}{2}}}dt\\\\\therefore x(t)=\frac{5}{0.5}\int t^{-0.5}dt\\\\x(t)=10\sqrt{t}+c$

To find the constant we note that at t=1 the particle is at x=12 Thus applying values in the above equation we get

$c=12-10\\\therefore c=2\\\\x(t)=10\sqrt{t}+2$

6 0

3 years ago