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ludmilkaskok [199]

2 years ago

14

A ball is rolling across the floor at a constant speed. What will happen to the ball if it is exposed to an unbalanced force in

the same direction that it is moving?
A. It will immediately stop.
B. Its speed will increase.
C. Its speed will decrease.
D. Its motion will be unchanged.

2 answers:

KATRIN_1 [288]2 years ago

7 0

Answer:

B. it's speed will increase

lions [1.4K]2 years ago

4 0

Explanation:

B. Its speed will increase.

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An object with charge q=−6.00×10−9C is placed in a region of uniform electric field and is released from rest at point A. After

shtirl [24]

Answer:

its a

Explanation:

used google

8 0

3 years ago

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

monitta

Answer:

force F = 1.66 × $10^{-13}$ N

Explanation:

given data

proton and an electron = 865 nm

solution

we get here force that is express as

force F = k q1 q2 ÷ r² ......................1

put here value and we get

force F = 9 × $10^{9}$ × $\frac{1.6\times (10^{-19})^{2}}{865 \times (10^{-9})^{2}}$

force F = 1.66 × $10^{-13}$ N

4 0

3 years ago

un conductor de plata (p=1,6x10°-6ohm x m) tiene una seccion de 5x10°-6 m°2 y una longitud de 110 m . calcular la resistencia

Alexus [3.1K]

Responder:

35,2 ohm.

Explicación:

Dado:

La resistencia específica del conductor es,

La longitud del conductor es,

El área de la sección transversal del conductor es,

Sabemos que la resistencia de un conductor es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.

Por lo tanto, la resistencia se puede expresar como:

$R=\frac{\rho\times l}{A}$

Ahora, conecte los valores dados y resuelva para 'R'. Esto da,

$R=\frac{1.6\times 10^{-6}\ ohm\cdot m\times 110\ m}{5\times 10^{-6}\ m^2}\\\\R=35.2\ ohm$

Por lo tanto, la resistencia del conductor es de 35,2 ohm.

4 0

3 years ago

A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

Artist 52 [7]

Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

7 0

3 years ago

Turn the ignition switch to start and release the key immediately or you could destroy the______________.

vredina [299]

Starter

Explanation:

Turn the ignition switch to start and release the key immediately or you could destroy the starter.

The car starter is used to cause ignition in the internal combustion engine in order to fire the piston and cause mechanical motion. The starter is used to start the cyclic process of the internal combustion engine.

Once the engine starts by igniting the starter, it is best to release it.
The starter ensures that the spark plug is engaged and the motor is brought into work.
If the ignition is still engaged, the process continues repeatedly and it can damage the starter of the car.

learn more:

Automobile brainly.com/question/2599962

#learnwithBrainly

5 0

3 years ago