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Anvisha [2.4K]
3 years ago
7

Show that the units of g, previously given as N/kg, are also m/s^2.

Physics
1 answer:
cupoosta [38]3 years ago
8 0

Explanation:

N/ Kg .................(1)

where,

1N= 1( Kg× m)/ s² ( sub in 1)

so,

(Kg×m)/ ( kg × s²)

kg is removed with kg

= m/ s²

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A toy car is moving along with 0. 50 joules of kinetic energy. If its speed is doubled, then its new kinetic energy is.
antoniya [11.8K]

Answer:

2 Joules

Explanation:

Kinetic Energy = 1/2 of (Mass x Velocity^2)

7 0
2 years ago
The pupil of an eagle’s eye has a diameter of 6.0 mm. Two field mice are separated by 0.010 m. From a distance of 185 m, the eag
max2010maxim [7]

Answer:

time is 5.973826 sec

Explanation:

Given data

diameter D =  6.0 mm  6× 10^{-3} m

separated d =  0.010 m

distance (dis) 185 m

speed s = 16 m/s

wavelength = 550 nm = 550  10^{-9} m

to find out

How much time passes

solution

we know that for resolution  we use Rayleigh's Criterion i.e

θ = 1.22 wavelength  / diameter  =  separated / distance 1

we calculate distance 1 by put value wavelength, diameter and   separated

distance 1  = diameter × separated / 1.22 wavelength

distance 1  = 6× 10^{-3} × 0.010   / 1.22 × 550 × 10^{-9}

distance 1 = 89.418778

so time will be i.e = distance (dis) - distance 1  / speed

time = ( 185 -  89.418778) / 16

time = 5.973826 sec

time is 5.973826 sec

4 0
3 years ago
A 92.6 kg weight-watcher wishes to climb a mountain to work off the equivalent of a large piece of chocolate cake rated at 735 (
Kitty [74]

Answer:

349 m

Explanation:

Parameters given:

Mass of climber, m = 92.6 kg

Amount of food calories = 735

1 food calorie = 103 calories

735 food calories = 75705 calories

1 joule is equal to 0.239 calories. Therefore, 75705 calories will be 316749.72 joules.

Hence, this is the amount of work the climber must do work off the food he ate.

Work done is given as:

W = Force * distance

W = m * g * h

h = W/(m * g)

h = 316749.72/(92.6 * 9.8)

h = 349 m

4 0
3 years ago
If a bullet loses 1/nth of its velocity while passing through a plank,then no of planks required to stop the bullet is?
Novay_Z [31]
Based on the answer provided, it seems the writer wanted you to assume that the energy loss per plank is constant. This is not the same as the bullet losing <span><span>1/nth</span><span>1/nth</span></span><span> of its velocity per plank (however, the fact that the question does not mention this assumption arguably makes the question ambiguous).

</span><span>With this assumption, the energy loss becomes
</span><span>
ΔE = <span>1/2 </span>m<span>v2 </span>− <span>1/2 </span>m <span><span>(<span>v−<span>v/n</span></span>) </span><span>2
</span></span></span>
and the number of planks <span>NN</span><span> becomes
</span>
N = <span><span><span>1/2</span>m<span>v2 /</span></span><span>ΔE </span></span>= <span><span>n2/ </span><span>2n−1
</span></span>
Otherwise, if you assume that the bullet loses <span><span>1/<span>nth</span></span><span>1/<span>nth</span></span></span><span> of its velocity per plank, then the answer is </span><span><span>N=∞</span></span><span><span>

</span>
</span>
8 0
3 years ago
Is 468km per minute describing speed or velocity
Luden [163]
Velocity probably well that’s what i would put
6 0
3 years ago
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