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Veronika [31]
3 years ago
8

If the newton is the product of kilograms and meters/second2 what units comprise the pound?

Physics
1 answer:
Kobotan [32]3 years ago
6 0

Answer:

Pound is the product of slug and foot/square second.

Explanation:

We are given that

Force=1 N

1N=1kg\times ms^{-2}

We have to find the units comprise the pound.

Force=1 Pound

Mass=Slug

Acceleration=ft/s^2

Therefore,

1 pound=1 slug\times fts^{-2}

Therefore, we can write as 1 pound is equal to the product of slug and ft/square second.

Hence, pound is the product of slug and foot/square second.

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Explanation:

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3 years ago
What is electrical resistance and types of electrical resistance
bazaltina [42]

Answer:

Electrical Resistance is a measure of the opposition to current flow in an electrical circuit

Types: variable resistance and set resistance

Explanation:

4 0
3 years ago
Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus
charle [14.2K]

Answer:

The total energy of the composite system is 7.8 J.

Explanation:

Given that,

Height = 0.15 m

Radius of circular arc = 0.27 m

Suppose, the entire track is friction less. a bullet with a m₁ = 30 g mass is fired horizontally into a block of wood with m₂ = 5.29 kg mass. the acceleration of gravity is 9.8 m/s.

Calculate the total energy of the composite system at any time after the collision.

We need to calculate the total energy of the composite system

Total energy of the system at any time = Potential energy of the system at the stopping point

E=mgh+Mgh

E=(m+M)gh

Put the value in to the formula

E=(30\times10^{-3}+5.29)\times 9.8\times0.15

E=7.8\ J

Hence, The total energy of the composite system is 7.8 J.

8 0
4 years ago
20 kg object travels 28 meter and stops. coefficient friction= 0.085 how much work was done by friction?
Tresset [83]
Assuming it is on a horizontal surface:
friction = μR
R = 20g (g is gravity 9.81)
so Friction = 0.085 x 20g
Work done is force x distance 
so Work done = 0.085 x 20g x 28
 = 466.956 J

7 0
4 years ago
Determine the stopping distances for a car with an initial speed of 88 km/h and human reaction time of 2.0 s for the following a
seropon [69]

Explanation:

Given that,

Initial speed of the car, u = 88 km/h = 24.44 m/s

Reaction time, t = 2 s

Distance covered during this time, d=24.44\times 2=48.88\ m

(a) Acceleration, a=-4\ m/s^2

We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{-(24.44)^2}{2\times -4}

s = 74.66 meters

s = 74.66 + 48.88 = 123.54 meters

(b) Acceleration, a=-8\ m/s^2

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{-(24.44)^2}{2\times -8}

s = 37.33 meters

s = 37.33 + 48.88 = 86.21 meters

Hence, this is the required solution.

4 0
4 years ago
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