All categories

2 years ago

A 4n force is applied to an object with a mass of 2kg what is the accelertion of the object

Physics

1 answer:

Pavel [41]2 years ago

5 0

As per Newton's law we know that

F = ma

here we know that

F = 4 N

m = 2 kg

so from this equation we will have

4 = 2 a

a = 4/2 = 2 m/s^2

so the acceleration will be 2 m/s^2

You might be interested in

A hydraulic turbine is used to generate power by using the water in a dam. The elevation difference between the free surfaces up

Serjik [45]

Answer:

0.906

Explanation:

Let g = 9.81 m/s2. We can calculate the rate of change in potential energy when m = 201kg of water is falling down a distance of h = 131m per second

$\dot{E_p} = \dot{m}gh = 201*9.81*131 = 258307 J/s (W) = 258.307 kW$

So the efficiency of the water turbine is the ratio of output power over input power:

$\frac{234}{258.307} = 0.906$

3 0

3 years ago

True/False: If an object stays still, but the observer moves, the Doppler Effect is still observered

Vladimir79 [104]

Well, Harry, what you said is not necessarily true the way you said it.
But we know what you mean, and what you meant to say is true.

The Doppler effect is observed if there is relative radial motion
between an object and an observer AND if the object happens
to be putting out sound or light in the observer's direction.

3 0

3 years ago

Read 2 more answers

A man pulled a 13.0 kg object 11.8 cm vertically with his teeth. (a) How much work (in J) was done on the object by the man in t

jonny [76]

Answer:

(a)The work done by the man is -15.03J.

(b)The force exerted on the object is 127.4N.

Explanation:

Mass of the object pulled by the man is -13kg

Object is lifted 11.8 cm vertical with his teeth it means (displacement = +11.8cm = +0.118m)

Acceleration due to gravity is $9.8 \mathrm{m} / \mathrm{s}^{2}$

(a) Calculating the work done:

Work done = mgh

Where "m" is mass of an object, "g" is acceleration due to gravity and "h" is the displacement.

$\text { Work }=-13 \times 9.8 \times(+0.118 \mathrm{m})$

$\text { Work }=-15.03 \mathrm{J}$

The work done by the man is -15.03J.

(b) Calculating the force:

Probably the man and the object are close to the exterior of the earth. If the rigidity required to maintained the object of consistent velocity interior the gravitational field of the earth is $\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}$

Thus the weight of the object is balanced by the force of the man's teeth on the object. That is

F = mg

$\mathrm{F}=13 \times 9.8$

F = 127.4N

The force exerted on the object is 127.4N.

4 0

2 years ago

The upward velocity of a 2540kg rocket is v(t)=At + Bt2. At t=0 a=1.50m/s2. The rocket takes off and one second afterwards v=2.0

Alexeev081 [22]

Answer:

The value of A is 1.5m/s^2 and B is 0.5m/s^³

Explanation:

The mass of the rocket = 2540 kg.

Given velocity, v(t)=At + Bt^2

Given t =0

a= 1.50 m/s^2

Now, velocity V(t) = A*t + B*t²

If, V(0) = 0, V(1) = 2

a(t) = dV/dt = A+2B × t

a(0) = 1.5m/s^²

1.5m/s^² = A + 2B × 0

A = 1.5m/s^2

now,

V(1) = 2 = A× 1 + B× 1^²

1.5× 1 +B× 1 = 2m/s

B = 2-1.5

B = 0.5m/s^³

Now Check V(t) = A× t + B × t^²

So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² × 1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s

Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)

7 0

3 years ago

The net force acting on an object in the horizontal direction is 25 newtons, and the force acting in the vertical direction is 1

Zina [86]

The angle is found by taking the arctangent of the vertical angle divided by the horizontal angle
θ = arctan(18/25) = 35.75°
36° is the closest answer.

4 0

3 years ago