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2 years ago

When acted upon by an unbalanced force, an object at rest will do what?

Physics

1 answer:

Firdavs [7]2 years ago

5 0

Answer:

When unbalanced forces act on an object at rest, the object will move. In the two examples mentioned earlier, the net force on the object is greater than zero. Unbalanced forces produced change in motion (acceleration) and the receivers of the forces - the piano and the rope - moved.

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A constant magnetic field passes through a single rectangular loop whose dimensions are 0.35 m × 0.55 m. The magnetic field has

Pani-rosa [81]

Answer:

Part a)

$EMF = 0.38 V$

Part b)

$\frac{dA}{dt} = 0.43 m^2/s$

Explanation:

Part a)

Initial value of magnetic flux is given as

$\phi_1 = BAcos\theta$

$\phi_1 = (2.1)(0.35 \times 0.55) cos65$

so we have

$\phi_1 = 0.17 Wb$

Final flux through the loop is given as

$\phi_2 = 0$

now EMF is given as

$EMF = \frac{\phi_1 - \phi_2}{\Delta t}$

$EMF = \frac{0.17 - 0}{0.45}$

$EMF = 0.38 V$

Part b)

If magnetic field is constant while Area is changing

So EMF is given as

$E = Bcos65 \times \frac{dA}{dt}$

$0.38 = 2.1 cos65(\frac{dA}{dt})$

$\frac{dA}{dt} = 0.43 m^2/s$

5 0

2 years ago

a car travelling 90 km/hr is 100m behind a truck travelling 75 km/hr. Assuming both vehicles moving at constant velocity,calcula

nika2105 [10]

90t - .100 = 75t

90t - 75t = 0.100

15t = 0.100

t = 0.0067

<u>t = 0.0067 hours or 24.12 seconds</u> (.0067 * 60 minutes * 60 seconds)

7 0

2 years ago

Physics!!! Please help!!!!!!!

Marianna [84]

Answer:

it is a very good morning amor de g the first paragraph of the first paragraph of the first paragraph of the first paragraph of the first paragraph of the first paragraph of 88th paragraph of the first paragraph

8 0

2 years ago

An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one fourth of its maximum height above its l

Murrr4er [49]

Answer:

v = 25.45 m/s

Explanation:

In order to calculate the initial speed of the object, you take into account the formula for the maximum height reaches by the object. Such a formula is given by:

$h_{max}=\frac{v_o^2}{g}$ (1)

vo: initial speed of the object = 18 m/s

g: gravitational acceleration = 9.8 m/s²

Furthermore you use the following formula for the final speed of the object:

$v^2=v_o^2-2gh$ (2)

h: height

You know that the speed of the object is 18m/s when it reaches one fourth of the maximum height. You use this information, and you replace the equation (1) in to the equation (2), as follow:

$v^2=v_o^2-2g(\frac{h_{max}}{4})=v_o^2-\frac{1}{2}g(\frac{v_o^2}{g})\\\\v^2=v_o^2-\frac{1}{2}v_o^2=\frac{1}{2}v_o^2$

Then, you solve the previous result for vo:

$v_o=\sqrt{2}v=\sqrt{2}(18m/s)=25.45\frac{m}{s}$

The initial speed of the object was 25.45 m/s

3 0

3 years ago

according to newtons second law of motion a larger force acting on an object causes a greater what of that object

inna [77]

A greater weight so the object has greater weight

3 0

3 years ago