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3 years ago

When there is an electric force acting on an object, the direction in which an object may accelerate has to do with

Physics

1 answer:

noname [10]3 years ago

6 0

Answer:

When an external non-zero net force acts on an object, the object accelerates in the direction of the net force. The magnitude of the acceleration is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object

hope this helps u

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Continuous signals are characterized as . signals that are broken up into binary code are characterized as . an am/fm radio is a

emmainna [20.7K]

Continuous signals are characterized as analog signals that are broken up into binary code are characterized as digital.

An am/fm radio is a good example of analog technology. A smart phone is a good example of digital technology.

<h3>What is a Digital technology?</h3>

These are devices that generate, store or process data while analog is a continuous signal that represents physical measurements.

The examples of analog and digital technology are mentioned above and is the most appropriate choice.

Read more about Analog and digital technology here brainly.com/question/26307469

4 0

2 years ago

Which car has the most kinetic energy?

Andru [333]

50kg driving at 30kph because it’s the heaviest yet fastest!

8 0

3 years ago

A basketball star covers 2.70 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled pr

Likurg_2 [28]

Answer:

Part a)

$T = 0.81 s$

Part b)

$v_x = 3.33 m/s$

Part c)

$v_y = 3.91 m/s$

Part d)

$\theta = 49.55 degree$

Part e)

$T = 1.11 s$

Explanation:

Part a)

initial vertical position = 1.02 m

maximum height = 1.80 m

$\Delta y = 1.80 - 1.02$

$\Delta y = 0.78 m$

$v_f^2 - v_y^2 = 2a \Delta y$

$0 - v_y^2 = 2(-9.81)(0.78)$

$v_y = 3.91 m/s$

time taken by it to reach this height

$v_y = v_i + at$

$0 = 3.91 - 9.81 t_1$

$t_1 = 0.39 s$

Now when it again touch the ground then its speed is given as

$v_f^2 - v_y^2 = 2a \Delta y$

$v_f^2 - 0 = 2(9.81)(1.80 - 0.95)$

$v_y = 4.08 m/s$

time taken by it to reach this height

$4.08 = v_i + at$

$4.08 = 0 + 9.81 t_2$

$t_2 = 0.42 s$

$T = t_1 + t_2$

$T = 0.81 s$

Part b)

Horizontal velocity

$v_x = \frac{x}{t}$

$v_x = \frac{2.70}{0.81}$

$v_x = 3.33 m/s$

Part c)

vertical velocity is the intial y direction velocity

$v_y = 3.91 m/s$

Part d)

Take off angle is given as

$tan\theta = \frac{3.91}{3.33}$

$\theta = 49.55 degree$

Part e)

initial vertical position = 1.20 m

maximum height = 2.50 m

$\Delta y = 2.50 - 1.20$

$\Delta y = 1.30 m$

$v_f^2 - v_y^2 = 2a \Delta y$

$0 - v_y^2 = 2(-9.81)(1.30)$

$v_y = 5.05 m/s$

time taken by it to reach this height

$v_y = v_i + at$

$0 = 5.05 - 9.81 t_1$

$t_1 = 0.51 s$

Now when it again touch the ground then its speed is given as

$v_f^2 - v_y^2 = 2a \Delta y$

$v_f^2 - 0 = 2(9.81)(2.50 - 0.72)$

$v_y = 5.9 m/s$

time taken by it to reach this height

$5.9 = v_i + at$

$5.9 = 0 + 9.81 t_2$

$t_2 = 0.60 s$

$T = t_1 + t_2$

$T = 1.11 s$

5 0

2 years ago

Salt is dissolved in a flask of tap water. Distilling the mixture causes the salt to separate from the water. Which type of ener

Igoryamba

Distillation, for the water to be seperated it must be heated to break the chemical bond.

8 0

3 years ago

Read 2 more answers

A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, a

pantera1 [17]

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

$120^2=130^2+50^2-2\times130\times50\cos\alpha$

$\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}$

$\alpha=\cos^{-1}(0.3846)$

$\alpha=67.38^{\circ}$

We need to calculate the angle β

Using cosine law

$50^2=130^2+120^2-2\times130\times120\cos\beta$

$\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}$

$\beta=\cos^{-1}(0.923)$

$\beta=22.63^{\circ}$

We need to calculate the force on 130 cm side

Using formula of force

$F_{130}=ILB\sin\theta$

$F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0$

$F_{130}=0$

We need to calculate the force on 120 cm side

Using formula of force

$F_{120}=ILB\sin\beta$

$F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63$

$F_{120}=0.1385\ N$

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

$F_{50}=ILB\sin\alpha$

$F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38$

$F_{50}=0.1385\ N$

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

8 0

3 years ago