Your classmate’s mass is 63 kg and the table weighs 500 N. Calculate the normal force on the table by the floor. Show your work!

Answers

Answer 1

Answer:

Answer:

$F_N=1234.8N$

Explanation:

Hello.

In this case, since the normal force is opposite to the total present weight, we can compute it by considering the mass of the classmate with the gravity to compute its weight, and the weight of the table:

$F_N=63kg*9.8m/s^2+500N\n\nF_N=617.4N+500N\n\nF_N=1234.8N$

Best regards.

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Answers

4.0 m/s2

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If 2050 J of heat are added to a 150 g object its temperature increases by 15°C.(a) What is the heat capacity of this object?
(b) What is the object's specific heat?

Answers

When an object gets heated by a temperature ΔT energy needed, E = mcΔT

Here energy is given E = 2050 J

Mass of object = 150 g

Change in temperature ΔT = 15 $^0C$ = 15 K

a) Heat capacity of an object equal to the ratio of the heat added to (or removed from) an object to the resulting temperature change.

So heat capacity = E/ΔT = 2050/15 = 136.67 J/K

b) We have E = mcΔT

c = $(2050)/(150*10^(-3)*15) = 911.11 J/kgK$

So object's specific heat = 911.11 J/kgK

What is the force required to accelerate 1 kilogram of mass at 1 meter per second per second.1 Newton

1 pound

1 kilometer

1 gram

Answers

Answer:

it's answer is 1 newton

A spring oscillator is designed with a mass of 0.231 kg. It operates while immersed in a damping fluid, selected so that the oscillation amplitude will decrease to 1.00% of its initial value in 9.43 s. Find the required damping constant for the system.

Answers

Answer:

.487 s⁻¹

Explanation:

Let damping constant be τ . The equation of decreasing amplitude can be written as

A = A₀ $e^{-\tau t$

A / A₀ = $e^{-\tau t$

At t = 9.43 s , A / A₀ = .01

.01 = $[e^{-\tau*9.43$

ln.01 = - 9.43 τ

-4.6 = -9.43τ

τ = .487 s⁻¹

Answer:

0.05508 kg/sec

Explanation:

mass of the oscillator m= 0.231 Kg

amplitude of oscillation given by

$A=A_0e^(-It)$

Ao= maximum amplitude

t= time and 1.00% of its initial value in t= 9.43 s.

A= 0.01Ao

⇒0.01=e^(-I×9.43)

ln100= 9.43×l

l=0.4883

we know that l= c/2m

c= damping constant

c= 2ml

=2×0.231×0.4883

=0.05508 kg/sec

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