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Delicious77 [7]
2 years ago
14

Imamu’s car is travelling twice as fast as Mosi’s car. Mosi travels 200 km in 2 hours. How long does it take Imamu to travel 50

km?
Physics
1 answer:
soldi70 [24.7K]2 years ago
8 0

Answer:

0.25hr

Explanation:

Given parameters:

Distance  = 200km

Time = 2hrs

  Now, let us find the rate which is the speed;

    Speed  = \frac{distance}{time}  

 Insert parameters:

      Speed  = \frac{200}{2}   = 100km/hr

Since the speed of Imamu = 2 x speed of Mosi = 200km/hr

Time taken  = \frac{distance}{speed}  = \frac{50}{200}   = 0.25hr

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pav-90 [236]
To answer this question, we will use the law of conservation of momentum which states that:
(m1+m2)Vi = m1V1 + m2V2 where:
m1 is the mass of the woman = 50 kg
m2 is the mass of the cart = 10 kg
Vi is the initial velocity (of woman and cart combined) = 5 m/sec
V1 is the final velocity of the woman = 7 m/sec
V2 is the final velocity of the cart that we need to calculate

Substitute with the givens in the above equation to get the final velocity of the cart as follows:
(50+10)(5) = (50)(7) + (10)V2
10V2 = -50
V2 = -5 m/sec
Note that the negative sign indicates that the cart is moving in an opposite direction to the others.
4 0
3 years ago
You are driving to the grocery store at 20 m/s. You are 110m from an intersection when the traffic light turns red. Assume that
oksano4ka [1.4K]

As we know that reaction time will be

t = 0.50 s

so the distance moved by car in reaction time

d = vt

d = 20 \times 0.50

d = 10 m

now the distance remain after that from intersection point is given by

d = 110 - 10 = 100 m

So our distance from the intersection will be 100 m when we apply brakes

now this distance should be covered till the car will stop

so here we will have

v_f = 0

v_i = 20 m/s

now from kinematics equation we will have

v_f^2 - v_i^2 = 2 a d

0 - 20^2 = 2(a)100

a = \frac{-400}{200} = -2 m/s^2

so the acceleration required by brakes is -2 m/s/s

Now total time taken to stop the car after applying brakes will be given as

v_f - v_i = at

0 - 20 = -2 (t)

t = 10 s

total time to stop the car is given as

T = 10 s + 0.5 s = 10.5 s

3 0
2 years ago
You are in a tall building located near the equator. What happens to your tangential speed due to the earth's rotation as you ri
8_murik_8 [283]

Answer:increases

Explanation:

If we are going upward in an elevator from the ground floor to the top floor then it indicates that your distance from the center of the earth is increasing while the time period remains the same.

If the radial distance is increased then the tangential velocity of the object must be increased because the time period is the same.

This can be best explained by taking an example of a car moving in a circle of radius r. If radial is increased for the same period then the car has to travel at a higher velocity to make in time.              

3 0
3 years ago
A 45.0 kg girl is standing on a 159 kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,
Ganezh [65]

Answer:

(a) v_g_i=1.08\frac{m}{s}

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Initially, the girl and the plank are at rest. So, relative to the ice, we have:

0=m_gv_g_i+m_pv_p_i\\v_p_i=-\frac{m_gv_g_i}{m_p_i}(1)

(a) The velocity of the girl relative to the ice is:

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Here, v_g_p is the velocity of the girl relative to the plank and v_p_i is the velocity of the plank relative to the ice.

Replacing (1) in (2):

v_g_i=v_g_p-\frac{m_gv_g_i}{m_p_i}\\v_g_i+\frac{m_gv_g_i}{m_p_i}=v_g_p\\v_g_i(1+\frac{m_g}{m_p})=v_g_p\\v_g_i=\frac{v_g_p}{1+\frac{m_g}{m_p}}\\v_g_i=\frac{1.38\frac{m}{s}}{1+\frac{45kg}{159kg}}\\v_g_i=1.08\frac{m}{s}

(b) According to (2), the velocity of the plank relative to the surface of ice is:

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The negative sing indicates that the plank is moving to the left.

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this is how Florine appears in the periodic table. which is one best piece of information that "9" gives about an atom of florin
Marianna [84]

Answer:

9 is the number of protons in the fluorine atom (atomic number)

Explanation:

3 0
3 years ago
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