A large 10.kg medicine ball is caught by a 70.kg student on the track team. If the ball was moving at 4.0 m/s, how fast will the
2 answers:
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Answer:
≅50°
Explanation:
We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:
Δx=V₀t+at²/2
And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:
Δx=(V₀cosθ)t+at²/2
Now luckily we are given everything we need to solve (or you found the info before posting here):
- Δx=760 m
- V₀=87 m/s
- t=13.6 s
- a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!
With that we can plug the values in to get:
Answer:
(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz
(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz
Explanation:
(a) The fundamental, f₁, frequency is given as follows;
Where;
T = The tension in the string
μ = The linear density of the string
L = The length of the string
f₁ = The fundamental frequency = 560 Hz
If the tension in the string is increased by 15%, we will have;
Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz
(b) When the string length is decreased by one-third, we have;
The new length of the string, = 2/3·L
The value of the fundamental frequency will then be given as follows;
When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.
<h2>Δt = 71.67 °C</h2>
The temperature change of water is equal to 71.67 °C
<h3>Explanation:</h3>
Given:
Amount of transferred energy = 30,000 K J
Mass of water = 100 ml
Initial temperature = 20°C
To find the change in temperature of water.
Formula for Heat capacity is given by
Q = m×c×Δt ........................................(1)
where:
Q = Heat capacity of the substance (in J)
m=mass of the substance being heated in grams(g)
c = the specific heat of the substance in J/(g.°C)
Δt = Change in temperature (in °C)
Δt = (Final temperature - Initial temperature) = T(f) - T(i)
Q = 30,000 J
Mass of water = m = 100 ml
1 ml = 1 g ................................................(2)
Therefore m = 100 ml = 100 g
Specific heat of water is c = 4.186 J /g.
Δt = ?
Substituting these in equation (1), we get
Q = m×c×Δt
Rearranging the terms for Δt,
Δt =
Δt =
Δt = 71.67 °C