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podryga [215]
3 years ago
9

Two balls are thrown off a building with the same speed, one straight up and one at a 45° angle. Which statement is true if air

resistance can be ignored?
a. Both hit the ground at the same time.
b. The one thrown at an angle hits the ground with a higher speed.
c. Both hit the ground with the same speed.
d. The one thrown at an angle hits the ground with a lower speed.
Physics
1 answer:
Sedaia [141]3 years ago
8 0

Answer:

c. Both hit the ground with the same speed.

Explanation:

Given that speed of the both balls are same.

We know that kinetic energy given as

KE=\dfrac{1}{2}mv^2

v=Speed  ,  m= mass

If mass of the the balls are same then their kinetic energy will be same.

The potential energy PE

PE= m g h

h =height

Given that h is same ,therefore their potential energy will be same.

But the time taken by both the balls will be different.

Therefore answer is c.

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Is there information in the previous question which relates to this one?
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A particle travels clockwise on a circular path of diameter​ R, monitored by a sensor on the circle at point​ P; the other endpo
kotykmax [81]

We make a graphic of this problem to define the angle.

The angle we can calculate through triangle relation, that is,

sin\theta = \frac{c}{QP}\\sin\theta = \frac{c}{R}\\\theta=sin^{-1}\frac{c}{R}

With this function we should only calculate the derivate in function of c

\frac{d\theta}{dc} = \frac{1}{\sqrt{1-\frac{c^2}{R^2}}}(\frac{c}{R})'\\\frac{d\theta}{dc} = \frac{1}{\sqrt{R^2-c^2}}

That is the rate of change of \theta.

b) At this point we need only make a substitution of 0 for c in the equation previously found.

\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{\sqrt{R^2-0}}\\\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{R}

Hence we have finally the rate of change when c=0.

6 0
3 years ago
According to Boyle’s Law, when pressure is increased...
klasskru [66]

Boyle's law states that when you "shrink"(change its size) a container the pressure increases so the best answer would be D-volume is decreased.

4 0
3 years ago
You have a stopped pipe of adjustable length close to a taut 62.0 cmcm, 7.25 gg wire under a tension of 4710 NN. You want to adj
Oksana_A [137]

Answer:

6 cm long

Explanation:

F = 4110N

Vo(speed of sound) = 344m/s

Mass = 7.25g = 0.00725kg

L = 62.0cm = 0.62m

Speed of a wave in string is

V = √(F / μ)

V = speed of the wave

F = force of tension acting on the string

μ = mass per unit density

F(n) = n (v / 2L)

L = string length

μ = mass / length

μ = 0.00725 / 0.62

μ = 0.0116 ≅ 0.0117kg/m

V = √(F / μ)

V = √(4110 / 0.0117)

v = 592.69m/s

Second overtone n = 3 since it's the third harmonic

F(n) = n * (v / 2L)

F₃ = 3 * [592.69 / (2 * 0.62)

F₃ = 1778.07 / 1.24 = 1433.927Hz

The frequency for standing wave in a stopped pipe

f = n (v / 4L)

Since it's the first fundamental, n = 1

1433.93 = 344 / 4L

4L = 344 / 1433.93

4L = 0.2399

L = 0.0599

L = 0.06cm

L = 6cm

The pipe should be 6 cm long

6 0
3 years ago
On average, both arms and hands together account for 13% of a person's mass, while the head is 7.0% and the trunk and legs accou
Feliz [49]

Answer:

Angular velocity, N_f = 242.36 rpm

Explanation:

The mass of the skater, M = 74.0 kg

Mass of each arm, m_{a} = 0.13 * \frac{M}{2} ( since it is 13% of the whole body and each arm is considered)

m_{a} = 0.13 * 37\\m_a = 4.81 kg

Mass of the trunk, m_{t} = M - 2m_{a}

m_t = 74 - 2(4.81)\\m_{t} = 64.38 kg

Total moment of Inertia = (Moment of inertia of the arms) + (Moment of inertia of the trunks)

(I_{T} )_i = 2(\frac{m_{a}L^2 }{12} + m_a(0.5L + R)^2) + 0.5 m_t R^2

(I_{T} )_i = 2(\frac{4.81 * 0.7^2 }{12} + 4.81(0.5*0.7 + 0.175)^2) + 0.5 *64.38* 0.175^2\\(I_{T} )_i = 3.052 + 0.986\\(I_{T} )_i = 4.038 kgm^2

The final moment of inertia of the person:

(I_{T} )_f = \frac{1}{2} MR^{2} \\(I_{T} )_f = \frac{1}{2} * 74*0.175^{2}\\(I_{T} )_f = 1.133 kg.m^2

According to the principle of conservation of angular momentum:

(I_{T} )_i w_{i} = (I_{T} )_f w_{f}\\w_{i} = 68 rpm = (2\pi * 68)/60 = 7.12 rad/s\\4.038 * 7.12 =1.133* w_{f}\\w_{f} = 25.38 rad/s\\w_{f} = \frac{2\pi N_f}{60} \\25.38 = \frac{2\pi N_f}{60}\\N_f = (25.38 * 60)/2\pi \\N_f = 242.36 rpm

3 0
2 years ago
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