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SCORPION-xisa [38]
3 years ago
8

A scalar quantity has

Physics
1 answer:
alexandr402 [8]3 years ago
3 0

A scalar quantity is described by a magnitude only. It includes speed, time, mass, and volume. There is no direction involved. Therefore, your answer is C- Magnitude but no direction.

I hope I've helped! :)

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Burka [1]
Their cognitive skills and their ability to learn
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A mass weight of 120N is hung from two strings. what is the tension?
kramer
The weight should be shared between the two string equally. Therefore, tension in each string, T is;

T = 120 N/2 = 60 N
7 0
3 years ago
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(6) A 75 kg human total footprint area is 0.05 m2 when wearing winter boots. Suppose that you want to walk on snow that can at m
fredd [130]

Answer:

0.25m²

Explanation:

We know that the summation of forces in the vertical direction is zero

So

PA-mg=0

A=mg/p

So

Substituting

A= 75* 9.8/3*10^-3

=0.25m² which is the total shoe area

3 0
3 years ago
How much power is needed to lift a 20-kg object to a height of 4.0 m in 2.5 seconds?
4vir4ik [10]

Answer:

313.92w

Explanation:

Formula for power:

P=W/∆t = Fv

Givens:

m=20kg

∆y=4.0m

∆t=2.5s

a=9.81m/s²

In order to find power, we first need to solve for work.

W=Fd (force*displacement), f=mg

W=mg∆y

W=(20kg)(9.81m/s²)(4.0m)

W=784.8J

P=W/∆t

P=784.8J/2.5s

P=313.92 watts

5 0
3 years ago
An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu
Ilia_Sergeevich [38]

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

\alpha = \frac{\omega_f - \omega_i}{t}

where we have

\alpha = -80.0 rad/s^2 is the angular acceleration (negative since the rotor is slowing down)

\omega_f is the final angular speed

\omega_i = 2000 rad/s is the initial angular speed

t = 10.0 s is the time interval

Solving for \omega_f, we find the final angular speed after 10.0 s:

\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

\alpha = \frac{\omega_f - \omega_i}{t}

If we re-arrange it for t, we get:

t = \frac{\omega_f - \omega_i}{\alpha}

where here we have

\omega_i = 2000 rad/s is the initial angular speed

\omega_f=0 is the final angular speed

\alpha = -80.0 rad/s^2 is the angular acceleration

Solving the equation,

t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s

6 0
3 years ago
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