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2 years ago

A constant force of 8.0 N is exerted for 4.0 s on a 16-kg object initially at rest. The change in speed of this object will be:

Physics

1 answer:

AfilCa [17]2 years ago

8 0

The change in speed of this object is 3m/s

According to Newton's second law;

F = ma

F = mv/t

Given the following parameters

Force F = 8.0N

mass m = 16kg

time t = 4.0s

Required

speed v

Substitute the given parameters into the formula

v = Ft/m

v = 8 * 6/16

v = 48/16

v = 3m/s

Hence the change in speed of this object is 3m/s

Learn more here: brainly.com/question/19072061

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Lose electrons because they have a negative charge.

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2 years ago

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A 0.683 kg mass moves in SHM at the end of a spring. It takes 1.41 s to move from the position with the spring fully extended to

dsp73

Answer:

Spring constant of the spring will be equal to 9.255 N /m

Explanation:

We have given mass m = 0.683 kg

Time taken to complete one oscillation is given T = 1.41 sec

We have to find the spring constant of the spring

From spring mass system time period is equal to $T=2\pi \sqrt{\frac{m}{K}}$ , here m is mass and K is spring constant

So $1.41=2\times 3.14\sqrt{\frac{0.683}{K}}$

$0.2245=\sqrt{\frac{0.683}{K}}$

Squaring both side

$0.0504=\frac{0.4664}{K}$

$K=9.255N/m$

So spring constant of the spring will be equal to 9.255 N /m

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2 years ago

How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium

Minchanka [31]

The height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

<h3>Pressure and temperature at equilibrium </h3>

The relationship between pressure and temperature can be used to determine the height risen by the water.

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

where;

V₁ = AL
V₂ = A(L - y)
P₁ = Pa
P₂ = Pa + ρgh
T₁ = 20⁰C = 293 K
T₂ = 10⁰ C = 283 k

$\frac{PaAL}{T_1} = \frac{(P_a + \rho gh)A(L-y)}{T_2} \\\\\frac{PaL}{T_1} = \frac{(P_a + \rho gh)(L-y)}{T_2} \\\\L-y = \frac{PaLT_2}{T_1(P_a + \rho gh)} \\\\y = L (1 - \frac{PaT_2}{T_1(P_a + \rho gh)})\\\\y = 4.2(1 - \frac{101325 \times 283}{293(101325\ +\ 1000 \times 9.8 \times 100)} )\\\\y = 3.8 \ m$

Thus, the height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

The complete question is below:

A diving bell is a 4.2 m -tall cylinder closed at the upper end but open at the lower end. The temperature of the air in the bell is 20 °C. The bell is lowered into the ocean until its lower end is 100 m deep. The temperature at that depth is 10°C. How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium?

Learn more about thermal equilibrium here: brainly.com/question/9459470

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3 0

1 year ago

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit

Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

$M=8.7\cdot 10^{13}kg$ is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

$F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N$

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

$m\frac{v^2}{r}=G\frac{Mm}{r^2}$

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s$

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3 years ago

If a car travels to Philadelphia, a distance of 60 miles, in 1 hour, what is the average

IgorC [24]

Answer:60mph

Explanation:

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2 years ago