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stiv31 [10]
3 years ago
5

John McRighteous has a mass of 90kg. Calculate johns weight.

Physics
2 answers:
Natalija [7]3 years ago
3 0

Answer: W=M*gravity. 90 * 9.8=882N. That is the weight on Earth.

Explanation:

Calculating the weight is W=M*gravity. 90 * 9.8=882N. That is the weight on Earth.

forsale [732]3 years ago
3 0

Answer:

198.4 lbs

Explanation:

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A student moves a box across the floor by exerting 56.7 N of force and doing 195 J of
Paha777 [63]

Answer:

A. 3.4 m

Explanation:

Given the following data;

Force = 56.7N

Workdone = 195J

To find the distance

Workdone is given by the formula;

Workdone = force * distance

Making "distance" the subject of formula, we have;

Distance = \frac {workdone}{force}

Substituting into the equation, we have;

Distance = \frac {195}{56.7}

Distance = 3.4 meters.

8 0
3 years ago
What is the target heart rate for a 24 year old?
andriy [413]

Answer:

196

Explanation:

subtract 24 from 220 to get your answer.

6 0
3 years ago
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If distance between two charges increased 5 times then force between them
Marrrta [24]

decreased 5 times

Explanation: if the force increases 5 times between them would decrease 5 times

4 0
3 years ago
Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of
viktelen [127]

Answer:

W=2.76\times 10^{-6}\ J

Explanation:

Given that,

The distance between two spheres, r = 25 cm = 0.25 m

The capacitance, C = 26 pF = 26×10⁻¹² F

Charge, Q = 12 nC = 12 × 10⁻⁹ C

We need to find the work done in moving the charge. We know that, work done is given by :

U=\dfrac{Q^2}{2C}

Put all the values,

U=\dfrac{(12\times 10^{-9})^2}{2\times 26\times 10^{-12}}\\\\U=2.76\times 10^{-6}\ J

So, the work done is 2.76\times 10^{-6}\ J.

8 0
3 years ago
My Notes Ask Your Teacher 6. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclin
Tresset [83]

Answer:

Kf > Ka = Kb > Kc > Kd > Ke

Explanation:

We can apply

E₀ = E₁

where

E₀: Mechanical energy at the beginning of the motion (top of the incline)

E₁: Mechanical energy at the end (bottom of the incline)

then

K₀ + U₀ = K₁ + U₁

If v₀ = 0  ⇒  K₀

and  h₁ = 0   ⇒    U₁ = 0

we get

U₀ = K₁    

U₀ = m*g*h₀ = K₁

we apply the same equation in each case

a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J

d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J

e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J

f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J

finally, we can say that

Kf > Ka = Kb > Kc > Kd > Ke

8 0
3 years ago
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