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Norma-Jean [14]
3 years ago
11

If a ball leaves the ground with a vertical velocity of 5.46 m/s, how long does it take

Physics
1 answer:
kifflom [539]3 years ago
8 0

Answer:

0.557 s

Explanation:

Given:

v₀ = 5.46 m/s

v = 0 m/s

a = -9.8 m/s²

Find: t

v = at + v₀

0 m/s = (-9.8 m/s²) t + 5.46 m/s

t = 0.557 s

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According to einstein's theory of simple relativity (_E + mc(2)_). BLANK is converted into BLANK.
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Answer:

energy is converted into mass

Explanation:

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2 years ago
Nikolas had an idea that he could use the compressed carbon dioxide in a fire extinguisher to propel him on his skateboard.
Vikentia [17]
The Newton’s law Nikolas would use to come up with this idea is the <span>Third law that states:

</span><span>When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.
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So, in this case, let's name the first Body A which is the skateboard and the second body B which is <span>the compressed carbon dioxide in a fire extinguisher. Then, as shown in the figure below, according to the Third law:

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3 years ago
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What force would be required to accelerate a 1,100kg car to 0.5 m/s2
OlgaM077 [116]

Answer:

the force required to accelerate a 1,100kg car is 550N

5 0
3 years ago
A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary
Snezhnost [94]

Explanation:

Let N_p\ and\ N_s are the number of turns in primary and secondary coil of the transformer such that,

\dfrac{N_p}{N_s}=\dfrac{1}{3}

A resistor R connected to the secondary dissipates a power P_s=100\ W

For a transformer, \dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}

V_s=(\dfrac{N_s}{N_p})V_p

V_s=3V_p...............(1)

The power dissipated through the secondary coil is :

P_s=\dfrac{V_s^2}{R}

100\ W=\dfrac{V_s^2}{R}

V_p^2=\dfrac{100R}{9}.............(2)

Let N_p'\ and\ N_s' are the new number of turns in primary and secondary coil of the transformer such that,

\dfrac{N_p'}{N_s'}=\dfrac{1}{24}

New voltage is :

V_s'=(\dfrac{N_s'}{N_p'})V_p'

V_s'=24V_p...............(3)

So, new power dissipated is P_s'

P_s'=\dfrac{V_s'^2}{R}

P_s'=\dfrac{(24V_p)^2}{R}

P_s'=24^2\times \dfrac{(V_p)^2}{R}

P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}

P_s'=6400\ Watts

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

3 0
3 years ago
Heat is transferred at a rate of 2 kW from a hot reservoir at 800 K to a cold reservoir at 300 K. Calculate the rate at which th
vekshin1

Answer:

0.00417 kW/K or 4.17 W/K

Second law is satisfied.

Explanation:

Parameters given:

Rate of heat transfer, Q = 2kW

Temperature of hot reservoir, Th = 800K

Temperature of cold reservoir, Tc = 300K

The rate of entropy change is given as:

ΔS = Q * [(1/Tc) - (1/Th)]

ΔS = 2 * (1/300 - 1/800)

ΔS = 2 * 0.002085

ΔS = 0.00417 kW/K or 4.17 W/K

Since ΔS is greater than 0, te the second law of thermodynamics is satisfied.

6 0
3 years ago
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