Answer:
k = 45.95 N/m
Explanation:
First, we will find the launch speed of the ball by using the formula for the horizontal range of the projectile.

where,
Vo = Launch Speed = ?
R = Horizontal Range = 5.3 m
θ = Launch Angle = 35°
Therefore,

v₀² = 55.33 m²/s²
Now, we know that the kinetic energy gain of ball is equal to the potential energy stored by spring:

where,
k = spring constant = ?
x = compression = 17 cm = 0.17 m
m = mass of ball = 24 g = 0.024 kg
Therefore,

<u>k = 45.95 N/m</u>
Answer:
A
Explanation:
60 + 30 = 90
90 divided by 3 = 30
30 divided by 60 = 0.5
so your answer is 0.5 m/s
Answer:
The work done on the wagon is 37 joules.
Explanation:
Given that,
The force applied by Charlie to the right, F = 37.2 N
The force applied by Sara to the left, F' = 22.4 N
We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :



Work done on the wagon is given by the product of net force and displacement. It is given by :


W = 37 Joules
So, the work done on the wagon is 37 joules. Hence, this is the required solution.
C. The downward component of the projectile's velocity continually increases
Explanation:
The motion of a projectile consists of two independent motions:
- A uniform motion (with constant velocity) along the horizontal direction
- A uniformly accelerated motion, with constant acceleration (equal to the acceleration of gravity) in the downward direction
Here we want to study the downward component of the projectile's velocity. Since the vertical motion is a uniformly accelerated motion, the vertical velocity is given by:

where
u = 0 is the initial vertical velocity (zero since the projectile is fired horizontally)
downward is the acceleration of gravity
t is the time
So the equation becomes

This means that
C. The downward component of the projectile's velocity continually increases
Because every second, it increases by
in the downward direction.
Learn more about projectile motion:
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Answer:
<em>faster and at a higher luminosity and temperature.</em>
Explanation:
A protostar looks like a star but its core is not yet hot enough for fusion to take place. The luminosity comes exclusively from the heating of the protostar as it contracts. Protostars are usually surrounded by dust, which blocks the light that they emit, so they are difficult to observe in the visible spectrum.
A protostar becomes a main sequence star when its core temperature exceeds 10 million K. This is the temperature needed for hydrogen fusion to operate efficiently.
Stars above about 200 solar masses (Higher mass) generate power so furiously that gravity cannot contain their internal pressure. These stars blow themselves apart and do not exist for long if at all. A protostar with less than 0.08 solar masses never reaches the 10 million K temperature needed for efficient hydrogen fusion. These result in “failed stars” called brown dwarfs which radiate mainly in the infrared and look deep red in color. They are very dim and difficult to detect, but there might be many of them, and in fact they might outnumber other stars in the universe.
That is why higher mass protostars enter the main sequence at a <em>faster and at a higher luminosity and temperature.</em>