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Zepler [3.9K]
3 years ago
12

What are the horizontal and vertical velocities of a stunt bike that leaves a ramp at 100 km/hr and at an angle of 35 degrees?

Physics
1 answer:
Alex787 [66]3 years ago
4 0

Horizontal velocity: 81.9 km/h

Vertical velocity: 57.4 km/h

Explanation:

We can solve this problem by resolving the velocity vector into its component along the horizontal and vertical direction.

The horizontal velocity of the stunt bike is given by:

v_x = v cos \theta

where

v = 100 km/h is the magnitude of the velocity

\theta=35^{\circ} is the angle of projection

Substituting, we find

v_x = (100)(cos 35^{\circ})=81.9 km/h

The vertical velocity instead is given by

v_y = v sin \theta

where

v=100 km/h

\theta=35^{\circ}

Substituting,

v_y = (100)(sin 35^{\circ})=57.4 km/h

Learn more about vector components:

brainly.com/question/2678571

#LearnwithBrainly

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What is Latent heat and also give types.<br>​
dmitriy555 [2]

Answer:

Latent heat is energy released or absorbed, by a body or a thermodynamic system, during a constant-temperature process. Two common forms of latent heat are latent heat of fusion (melting) and latent heat of vaporization (boiling).

Explanation:

8 0
2 years ago
A player kicks a soccer ball in a high arc toward the opponent's goal. At the highest point in its trajectory
____ [38]

At the highest point in its trajectory, the ball's acceleration is zero but its velocity is not zero.

<h3>What's the velocity of the ball at the highest point of the trajectory?</h3>
  • At the highest point, the ball doesn't go more high. So its vertical velocity is zero.
  • However, the ball moves horizontal, so its horizontal component of velocity is non - zero i.e. u×cosθ.
  • u= initial velocity, θ= angle of projection

<h3>What's the acceleration of the ball at the highest point of projectile?</h3>
  • During the whole projectile motion, the earth exerts the gravitational force with a acceleration of gravity along vertical direction.
  • But as there's no acceleration along vertical direction, so the acceleration along vertical direction is zero.

Thus, we can conclude that the acceleration is zero and velocity is non-zero at the highest point projectile motion.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Player kicks a soccer ball in a high arc toward the opponent's goal. At the highest point in its trajectory

A- neither the ball's velocity nor its acceleration are zero.

B- the ball's acceleration points upward.

C- the ball's acceleration is zero but its velocity is not zero.

D- the ball's velocity points downward.

Learn more about the projectile motion here:

brainly.com/question/24216590

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7 0
2 years ago
A 0.10-kg ball, traveling horizontally at 25 m/s, strikes a wall and rebounds at 19 m/s. What is the 7) magnitude of the change
IRISSAK [1]

Answer:

Change in momentum will be -4.4 kgm/sec

So option (A) is correct option

Explanation:

Mass of the ball is given m = 0.10 kg

Initial velocity of ball v_1=25m/sec

And velocity after rebound v_2=-19m/sec

We have to find the change in momentum

So change in momentum is equal to =m(v_2-v_1)=0.1\times (-19-25)=-4.4kgm/sec ( here negative sign shows only direction )

So option (A) will be correct answer

5 0
3 years ago
Select the correct term to complete each sentence. If you the wavelength, the electromagnetic radiation energy will double. The
Andrews [41]

Answer:

A) If you halve the wavelength, the electromagnetic radiation energy will double.

B) The energy of the electromagnetic radiation will halve if you halve the wavenumber.

C) When the frequency of the light is doubled, its energy will double.

Explanation:

The function for the light frequency is given as

The energy supplied to each electron is doubled by halving the wavelength, nearly doubling its kinetic energy by two after it is free from the metal. It is important to remember that for a given period of time, the number of electrons ejected will remain constant.

Cheers

5 0
2 years ago
At the Equator near Earth’s surface, the magnetic field is approximately 82.2 µT northward and the electric field is about 143 N
IrinaK [193]

Answer:

Magnitude of gravitational force of the electron= 3.74×10^12N

Explanation:

Felectron= Fgravitational

Therefore:

Felectron/Fgravitational = kq^2/r^2 ×(r^2/Gm^2)

= kq^2/Gm^2

Where G= gravitational constant

m = mass of electron=9.1×10^-31

K= 8.9x10^9

q= 1.6×10^-19

Magnitude of gravitational force= (8.9×10^9)×(1.6×10^-19)^2/(6.7×10^-11)×(9.1×10^-31)

=( 2.2784×10^-28) / ( 6.097×10^-41)

= 3.74×10^12N

7 0
3 years ago
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