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3 years ago

James Bond is trying to escape his enemy on a speedboat but

Physics

1 answer:

SVEN [57.7K]3 years ago

3 0

Answer:

100 m/s

Explanation:

Mass the mass of Bond's boat is m₁. His enemy's boat is twice the mass of Bond's i.e. m₂ = 2 m₁

Initial speed of Bond's boat is 0 as it won't start and remains stationary in the water. The initial speed of enemy's boat is 50 m/s. After the collision, enemy boat is completely stationary. Let v₁ is speed of bond's boat.

It is the concept of the conservation of momentum. It remains conserved. So,

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

Putting all the values, we get :

$0+(2m_1)50=m_1v_1+(2m_2)(0)\\\\100m_1=m_1v_1\\\\v_1=100\ m/s$

So, Bond's boat is moving with a speed of 100 m/s after the collision.

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Answer:

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Explanation:

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2 years ago

Which of the following events happens first in the development of a new organism? A. A sperm fertilizes an egg to form a zygote.

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A cannon sends a projectile towards a target a distance 1420 m away. The initial velocity makes an angle 35◦ with the horizontal

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Answer:

$v_{o}=141.51m/s$

Explanation:

From the exercise we know the final x distance, the angle which the projectile is being released and acceleration of gravity

$x=1420m\\g=-9.8m/s^{2}$

From the equation of x-position we know that

$x=v_{ox}t=v_{o}cos(35)t$

Solving for $v_{o}$

$v_{o}=\frac{x}{tcos(35)} =\frac{1420m}{tcos(35)}$ (1)

Now, if we analyze the equation of y-position we got

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

At the end of the motion y=0

$0=v_{o}sin(35)t+\frac{1}{2}gt^{2}$

Knowing the equation for $v_{o}$ in (1)

$0=\frac{1420}{tcos(35)}tsin(35)-\frac{1}{2}(9.8)t^{2}$

$\frac{1}{2}(9.8)t^{2}=1420tan(35)$

Solving for t

$t=\sqrt{\frac{2(1420tan(35))}{9.8} } =14.25s$

Now, we can solve (1)

$v_{o}=\frac{1420m}{(14.25s)cos(35)}=141.51m/s$

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Gravity on the surface of the moon is only 1/6 as strong as gravity on the Earth. What is the weight of a 19 kg object on the Ea

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The weight of anything in any place is

(mass of the thing) x (acceleration of gravity in that place).

-- On Earth, the acceleration of gravity is about 9.807 m/s²

Weight of 19 kg of mass is (19 kg) x (9.807 m/s²) = 186.3 newtons

-- On the Moon, the acceleration of gravity is about 1.623 m/s²

Weight of the same 19 kg of mass is (19 kg) x (1.623 m/s²) = 30.8 newtons

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3 years ago

A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period Δt = 1.50 s, the tire is stopped and spun in the

Andru [333]

Answer:

Δω = -5.4 rad/s

αav = -3.6 rad/s²

Explanation:

Given:

Initial angular velocity = ωi = 2.70 rad/s

Final angular velocity = ωf = -2.70 rad/s (negative sign is

due to the movement in opposite direction)

Change in time period = Δt = 1.50 s

Required:

Change in angular velocity = Δω = ?

Average angular acceleration = αav = ?

Solution:

Angular velocity (Δω):

Δω = ωf - ωi

Δω = -2.70 - 2.70

Δω = -5.4 rad/s.

Average angular acceleration (αav):

αav = Δω/Δt

αav = -5.4/1.50

αav = -3.6 rad/s²

Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.

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