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3 years ago

In the graph above, what is the instantaneous speed of the object after the first five seconds?

Physics

1 answer:

alexdok [17]3 years ago

8 0

Answer:

D) 25 m/s

Explanation:

In order to solve this problem we must use the following kinematics equation.

$v_{f} =v_{i} +(a*t)\\$

where:

Vf = final speed [m/s]

Vi = initial speed = 0

a = acceleration = 5[m/s^2]

t = time = 5[s]

After 5 seconds the acceleration is equal to 5 [m/s^2]

Now replacing the values in the equation:

Vf = 0 + (5*5)

Vf = 25[m/s]

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A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s.

kherson [118]

Answer:

a. The rock takes 2.02 seconds to hit the ground

b. The rock lands at 20,2 m from the base of the cliff

Explanation:

Horizontal motion occurs when an object is thrown horizontally with an initial speed v from a height h above the ground. When it happens, the object moves through a curved path determined by gravity until it hits the ground.

The time taken by the object to hit the ground is calculated by:

$\displaystyle t=\sqrt{\frac{2h}{g}}$

The range is defined as the maximum horizontal distance traveled by the object and it can be calculated as follows:

$\displaystyle d=v.t$

The man is standing on the edge of the h=20 m cliff and throws a rock with a horizontal speed of v=10 m/s.

The time taken by the rock to reach the ground is:

$\displaystyle t=\sqrt{\frac{2*20}{9.8}}$

$\displaystyle t=\sqrt{4.0816}$

t = 2.02 s

The rock takes 2.02 seconds to hit the ground

The range is calculated now:

$\displaystyle d=10\cdot 2.02$

d = 20.2 m

The rock lands at 20,2 m from the base of the cliff

5 0

3 years ago

An egg is thrown upward with a velocity of 4.5 m/s. How long will it take to reach it's maximum height?

Ray Of Light [21]

Answer:

0.45 seconds

Explanation:

Letting the value of g = 10 m/s/s

final velocity (v) = 0 m/s (since the egg will come to rest at the maximum height)

initial velocity(u) = 4.5 m/s

acceleration = -10 m/s/s (since the gravity is acting against the egg)

time = t seconds

From the first equation of motion:

v = u + at

0 = 4.5 + (-10)t

t = -4.5 / -10

t = 0.45 seconds

3 0

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

10. 1. Why is Newton's law of gravitation called universal law ?

Eduardwww [97]

Newtons law of gravitation is called the universal law of gravitation because it is applicable for all masses at all distances, independent of the medium.

6 0

2 years ago

Read 2 more answers

In September 2016 you purchased a 125 shares of Grainger stock for $224.84. It is now September 2019 and Grainger stock is curre

Temka [501]

Answer:

Gain in capital = $ 70.72

Explanation:

Given:

- The price of stocks when purchased P_o = $ 224.84

- The price of stocks when sold P_s = $ 295.56

Find:

what would be your capital gain (loss) on the sale, ignoring commissions

Solution:

- The capital gain or loss on the selling of stocks stems from the difference of buying and selling value of stocks. The original price of stock was P_o and the selling price would be P_s. The difference would be:

capital gain = P_s - P_o

capital gain = $295.56 - $224.84

capital gain = $ 70.72

- Hence, there would be a gain in capital if sold today for about $ 70.72.

6 0

3 years ago