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Airida [17]
3 years ago
12

The third one pls help

Physics
1 answer:
Nadya [2.5K]3 years ago
7 0

Answer:

20 ms¯¹

Explanation:

3. Determination of the final velocity

From the question given above, the following data were obtained:

Time (t) = 4 s

Acceleration (a) = 5 ms¯²

Initial velocity (u) = 0 ms¯¹

Final velocity (v) =?

Acceleration is simply defined as the change in velocity per unit time.

Mathematically, it can be expressed as:

Acceleration (a) = final velocity – Initial velocity / time

a = v – u / t

With the above formula, we can obtain the final velocity of the car as follow:

Time (t) = 4 s

Acceleration (a) = 5 ms¯²

Initial velocity (u) = 0 ms¯¹

Final velocity (v) =?

a = v – u / t

5 = v – 0 / 4

5 = v / 4

Cross multiply

v = 5 × 4

v = 20 ms¯¹

Thus, the final velocity of the car is 20 ms¯¹

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3 years ago
An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr
mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = 5 \ * (M \ _{sun})

where : M_{sun} = 1.99*10^{33} \ kg

Then; we have:

 m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear m_{ear} = 0.030 \ kg

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R -  d) \omega^2

\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)

The net force on the ear farther to the black hole is :

\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R +  d) \omega^2

\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

T = \frac{3GMm_{ear}d}{R^3}

T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}

T = 2073.9 N\\T = 2.07 KN

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0
3 years ago
For a top player, a tennis ball may leave the racket on the serve with a speed of 55 m/s (about 120 mi/h). If the ball has a mas
inn [45]

Answer:

Yes is large enough

Explanation:

We need to apply the second Newton's Law to find the solution.

We know that,

F= ma

And we know as well that

a= \frac{v}{t}

Replacing the aceleration value in the equation force we have,

F= \frac{mv}{t}

Substituting our values we have,

F= \frac{(0.060)(55)}{4*10^{-3}}

F=825N

The weight of the person is then,

W = mg

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<em>We can conclude that force on the ball is large to lift the ball</em>

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What happens if a mid-ocean ridge occurs on land
Gnesinka [82]

Answer:

Despite being such prominent feature on our planet, much of the mid-ocean ridge system remains a mystery. While we have mapped about half of the global mid-ocean ridge in high resolution, less than one percent of the mid-ocean ridge has been explored in detail using submersibles or remotely operated vehicles. so therefore we do not have enough information about them to know what will happen

Explanation:

A mid-ocean ridge or mid-oceanic ridge is an underwater mountain range, formed by plate tectonics. This uplifting of the ocean floor occurs when convection currents rise in the mantle beneath the oceanic crust and create magma where two tectonic plates meet at a divergent boundary. Mid-ocean ridges occur along divergent plate boundaries, where new ocean floor is created as the Earth’s tectonic plates spread apart. As the plates separate, molten rock rises to the seafloor, producing enormous volcanic eruptions of basalt. The speed of spreading affects the shape of a ridge  slower spreading rates result in steep, irregular topography while faster spreading rates produce much wider profiles and more gentle slopes.

4 0
3 years ago
A man strikes one end of a thin rod with a hammer. The speed of sound in the rod is 15 times the speed of sound in air. A woman,
lina2011 [118]

Answer:

44.1 m

Explanation:

<u>Given:</u>

  • V_a = speed of sound in air = 343 m/s
  • V_r = speed of sound in the rod = 15V_a
  • \Delta t = times interval between the hearing the sound twice = 0.12 s

<u>Assumptions:</u>

  • l = length of the rod
  • t = time taken by the sound to travel through the rod
  • T = time taken by the sound to travel to through air to the same point = t+\Delta t = t+0.12\ s

We know that the distance traveled by the sound in a particular medium is equal to the product of the speed of sound in that medium and the time taken.

For traveling sound through the rod, we have

l=V_r t\\\Rightarrow t = \dfrac{l}{V_r}..........eqn(1)

For traveling sound through the air to the women ear for traveling the same distance, we have

l=V_aT\\\Rightarrow l=V_a(t+0.12)\\\Rightarrow l=V_a(\dfrac{l}{V_r}+0.12)\,\,\,\,\,\,(\textrm{From eqn (1)})\\\Rightarrow l=V_a(\dfrac{l}{15V_a}+0.12)\\\Rightarrow l=\dfrac{l}{15}+0.12V_a\\\Rightarrow l-\dfrac{l}{15}=0.12V_a\\\Rightarrow \dfrac{14l}{15}=0.12V_a\\\Rightarrow l = \dfrac{15}{14}\times 0.12V_a\\\Rightarrow l = \dfrac{15}{14}\times 0.12\times 343\\\Rightarrow l = \dfrac{15}{14}\times 0.12\times 343\\\Rightarrow l = 44.1\ m

Hence, the length of the rod is 44.1 m.

4 0
3 years ago
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