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zzz [600]
2 years ago
6

A motorboat embarks on a trip, heading downstream in a river in which the current flows at a rate of 1.5m/s. After 30.0 minutes,

the boat has traveled a distance of 24.3 km downstream. How long will it take the boat to travel upstream to its original point of embarkation
Physics
1 answer:
Natali5045456 [20]2 years ago
3 0

Answer:

t=2413s

Explanation:

From the question we are told that:

Velocity v=1.5m/s

Time t=30min=>30*60=>1800

Distance d=24.3km

Generally the Newton's equation for Speed going down the stream is mathematically given by

 v + u = \frac{d}{t}

 1.5+v=frac{24300}{1800}

 v=12m/s

Therefore

 v + u = \frac{d}{t}

 t=\frac{24300}{12-1.5}

 t=2413s

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Ganezh [65]

At the center of a 50 m diameter circular ice rink, if a 77 kg skater traveling at 2.3 m/s and then collides with a 63 kg skates traveling at 3.7 m/s. This is how long it will take them to glide to the edge of the rink:

Speed after the collision= √{[77(2.3)77^2] + [63(3.7)^2]} / (77+63)=2.09 m/s

For them to be able to get to the edge which is 50 m away it will take them 23.9 seconds.

8 0
3 years ago
Mr Ndlovu works in centurion and works at Rosebank.It takes him 13minutes to get to work and he needs to be there at 08h15
VARVARA [1.3K]

Answer:

8:02 no less

Explanation:

8:15 - 13mins = 8:02

6 0
3 years ago
In a football game, a 90 kg receiver leaps straight up in the air to catch the 0.42 kg ball the quarterback threw to him at a vi
irinina [24]

To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where,

m_{1,2}= Mass of each object

v_{1,2} = Initial velocity of each object

v_f= Final Velocity

Since the receiver's body is static for the initial velocity we have that the equation would become

m_2v_2 = (m_1+m_2)v_f

(0.42)(21) = (90+0.42)v_f

v_f = 0.0975m/s

Therefore the velocity right after catching the ball is 0.0975m/s

8 0
3 years ago
A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v1=8.0 meters per second. At height
AleksandrR [38]

a) 5.0 m/s

This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:

E_i=\frac{1}{2}m_1v_1^2

where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.

When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:

E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh

where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find

\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s

b) 4.0 m/s

After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:

p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3

where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find

v_3 = \frac{m_1 v_2}{m_1 + m_2}=\frac{(60 kg)(5.0 m/s)}{60 kg+15 kg}=4 m/s

c) 2.8 m

We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:

E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J

While at the maximum height, the speed is zero, so all the mechanical energy is just potential energy:

E_f = (m_1 +m_2)gh_{max}

where h_max is the maximum height. Equalizing the two expressions (because the mechanical energy must be conserved) and solving for h_max, we find

E_i = (m_1+m_2)gh_{max}\\h_{max}=\frac{E_i}{(m_1+m_2)g}=\frac{2070 J}{(75 kg)(9.8 m/s^2)}=2.8 m

4 0
3 years ago
Any example of acceleration related to human body??? <br> answer quickly plz :)
Hitman42 [59]

Answer:

Acceleration stress, physiological changes that occur in the human body in motion as a result of rapid increase of speed. ... A force of 3 g, for example, is equivalent to an acceleration three times that of a body falling near Earth.

4 0
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