All categories

4 years ago

If a 3.5 kg object is accelerating at 0.80 /s^2, the net force F causing this motion is ____ N.

Physics

1 answer:

lara31 [8.8K]4 years ago

6 0

Answer:

<h3>The answer is option B</h3>

Explanation:

To find the force acting on an object we use the formula

<h3>force = mass × acceleration</h3>

From the question

mass = 3.5 kg

acceleration = 0.80 m/s²

We have

Force = 3.5 × 0.8

We have the final answer as

Hope this helps you

You might be interested in

A treasure map instructs you to walk (25.0+A) m south, followed by (18.0+B) m east, then 64.5 m north, and finally 28.5 m east.

adelina 88 [10]

Answer:

Explanation:

Check attachment for solutions

3 0

3 years ago

A box on a table has these forces acting on it

umka21 [38]

Answer:

greater than 10 n

Explanation:

8 0

3 years ago

A car mass 600kg starts from rest moving uniform acceleration 0.2 m/s^2 after 60 seconds collides with stationary pick up van of

galben [10]

Answer:

the phenomenon of the system supports the principle of conservation of momentum.

Explanation:

The law of conservation of momentum says that:

Initial Momentum = Final Momentum

So, first we calculate initial momentum of the system:

Initial Momentum = m₁u₁ + m₂u₂

where,

m₁ = mass of car = 600 kg

m₂ = mass of van = 400 kg

u₁ = Initial Speed of Car

For initial speed of car, we use:

Vf = Vi + at

Vf = 0 m/s + (0.2 m/s²)(60 s)

Vf = u₁ = 12 m/s

u₂ = Initial Speed of Van = 0 m/s

Therefore,

Initial Momentum = (600 kg)(12 m/s) + (400 kg)(0 m/s)

Initial Momentum = 7200 Ns --------------- equation (1)

Now, for the final momentum:

Final Momentum = m₁v₁ + m₂v₂

where,

v₁ = v₂ = Final Speed of Car and van (both are locked) = 7.2 m/s

Therefore,

Final Momentum = (600 kg)(7.2 m/s) + (400 kg)(7.2 m/s)

Final Momentum = 7200 Ns ------------- equation (2)

Comparing equation (1) and (2):

Initial momentum = Final Momentum

<u>Hence, the phenomenon of the system supports the principle of conservation of momentum.</u>

5 0

3 years ago

A wheel moves in the xy plane in such a way that the location of its center is given by the equations xo = 12t3 and yo = R = 2,

Stella [2.4K]

Answer:

the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is $P = 104.04 \hat{i} -314.432 \hat{j}$

Explanation:

The free-body diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;

the peripheral velocity that is directed downward $(-V_y)$ along the y-axis

the linear velocity $(V_x)$ that is directed along the x-axis

Now;

$V_x = \frac{d}{dt}(12t^3+2) = 36 t^2$

$V_x = 36(1.7)^2\\\\V_x = 104.04\ ft/s$

Also,

$-V_y = R* \omega$

where $\omega$ (angular velocity) = $\frac{d\theta}{dt} = \frac{d}{dt}(8t^4)$

$-V_y = 2*32t^3)\\\\\\-V_y = 2*32(1.7^3)\\\\-V_y = 314.432 \ ft/s$

∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is $P = 104.04 \hat{i} -314.432 \hat{j}$

3 0

4 years ago

A movable piston having a mass of 8.00 kg and a cross-sectional area of 5.00 cm2 traps 0.430 moles of an ideal gas in a vertical

svp [43]

We can assume the process to be adiabatic such that we can make use of the formula:
W = R (T2 - T1) / (γ - 1)
W = 8.314 (297 - 17) / (1.4 - 1)
W = 700 J/mole
multiplying the number of moles
W = 700 (0.43)
W = 301 J
The work done is 301 J.

5 0

4 years ago