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Goryan [66]
3 years ago
15

The Fitness Gram push-up test is a measure of

Physics
1 answer:
Morgarella [4.7K]3 years ago
6 0

Answer:

The answer is B.

Explanation:

I meant B. not C so sorry

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Marcy pulls a backpack on wheels down the 100 m hall. The 60N force is applied at an angle of 30° above the horizontal. How much
shutvik [7]

Work= Fcos∆×S

W=60N×cos 30⁰×100

W=60×0.866×100=5196.1J

6 0
3 years ago
The magnetic field at the centre of a toroid is 2.2-mT. If the toroid carries a current of 9.6 A and has 6.000 turns, what is th
ziro4ka [17]

Answer:

Radius, r = 0.00523 meters

Explanation:

It is given that,

Magnetic field, B=2\ mT=2.2\times 10^{-3}\ T

Current in the toroid, I = 9.6 A

Number of turns, N = 6

We need to find the radius of the toroid. The magnetic field at the center of the toroid is given by :                  

B=\dfrac{\mu_oNI}{2\pi r}

r=\dfrac{\mu_oNI}{2\pi B}  

r=\dfrac{4\pi \times 10^{-7}\times 6\times 9.6}{2.2\pi \times 2\times 10^{-3}}  

r = 0.00523 m

or

r=5.23\times 10^{-3}\ m

So, the radius of the toroid is 0.00523 meters. Hence, this is the required solution.

7 0
3 years ago
Read 2 more answers
The two pucks of equal mass did not move linearly (they came to a stop) after the collision due to the conservation of linear mo
weeeeeb [17]

Compared to the pucks given, the pair of pucks will rotate at the same rate.

Answer: Option A

<u>Explanation:</u>

The law of conservation of the angular momentum expresses that when no outer torque follows upon an article, no difference in angular momentum will happen.  At the point when an item is turning in a shut framework and no outside torques are applied to it, it will have no change in angular momentum.

The conservation of the angular momentum clarifies the angular quickening of an ice skater as she brings her arms and legs near the vertical rotate of revolution.  In the event, that the net torque is zero, at that point angular momentum is steady or saved.  

By twice the mass yet keeping the speeds unaltered, also twice the angular momentum's to the two-puck framework.  Be that as it may, we likewise double the moment of inertia. Since L=I \times \omega, the turning rate of the two-puck framework must stay unaltered.

4 0
3 years ago
A fire helicopter carries a 700-kg bucket of water at the end of a 20.0-m long cable. Flying back from a fire at a constant spee
rodikova [14]

Answer:

F = 50636.873 N

Explanation:

given,

bucket of water =  700-kg

length of cable = 20 m

Speed  = 40 m/s

angle of the cable = 38.0°

let air resistance be = F

tension in rope be = T

T cos 38° = m×g..................(1)

T sin 38^0= \dfrac{mv^2}{l} + F..........(2)

equation (1)/(2)

tan 38^0 =\dfrac{\dfrac{mv^2}{l} + F}{mg}

       0.781=\dfrac{\dfrac{700\times 40^2}{20} + F}{700\times 9.8}

           F = 50636.873 N

Hence the force exerted on the bucket is equal to F = 50636.873 N

5 0
3 years ago
A 13.0 kg wheel, essentially a thin hoop with radius 1.80 m, is rotating at 469 rev/min. It must be brought to a stop in 16.0 s.
Stella [2.4K]

Answer:

Explanation:

Given

mass of wheel m=13 kg

radius of wheel=1.8 m

N=469 rev/min

\omega =\frac{2\pi \times 469}{60}=49.11 rad/s

t=16 s

Angular deceleration in 16 s

\omega =\omega _0+\alpha \cdot t

\alpha =\frac{\omega }{t}=\frac{49.11}{16}=3.069 rad/s^2

Moment of Inertia I=mr^2=13\times 1.8^2=42.12 kg-m^2

Change in kinetic energy =Work done

Change in kinetic Energy=\frac{I\omega ^2}{2}-\frac{I\omega _0^2}{2}

\Delta KE=\frac{42.12\times 49.11^2}{2}=50,792.34 J

(a)Work done =50.79 kJ

(b)Average Power

P_{avg}=\frac{E}{t}=\frac{50.792}{16}=3.174 kW

7 0
3 years ago
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