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iragen [17]
2 years ago
13

A colony of troglodytes has been in a lengthy feud with

Physics
1 answer:
Troyanec [42]2 years ago
7 0

Answer:

h = 181.73 m

Explanation:

given,

distance between the neighbors = 42 m

speed of the rock rolling = 6.9 m/s

vertical velocity of ball = 0 m/s

height of the cliff = ?

time taken by the ball to travel 42 m

d = s x t  

s is the horizontal speed of the ball equal to 6.9 m/s

t =\dfrac{d}{s}

t =\dfrac{42}{6.9}

t = 6.09 s

same time will be taken by the ball to travel vertical distance

Using equation of motion

h = u t + \dfrac{1}{2}gt^2

h = 0+ \dfrac{1}{2}gt^2

h = \dfrac{1}{2}\times 9.8 \times 6.09^2

 h = 181.73 m

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The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal cont
ivann1987 [24]

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

ΔT is the change in temperature

The heat transferred by the  object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}

But heat capacity of object B is twice that of object A

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K

Therefore, the final temperature of both objects is 400 K.

4 0
2 years ago
A gas is collected from a radioactive material; upon inspection, the gas is identified as helium. the presence of the helium ind
Flura [38]
The presence of helium gas indicates the radioactive sample is most likely decaying by α-decay, or alpha decay. α-decay is the type of radioactive decay in which an atomic nucleus emits α particles. α particles are Helium nuclei. So the correct answer would be α-decay.
7 0
2 years ago
6. A wave has a frequency of 600 Hz and is traveling at 300 m/s. What is its<br> wavelength?
Luda [366]

Answer:

0.5m

Explanation:

v=f×lamda

v is 300m/s, f is 600Hz, lamda is ?

lamda=v/f

lamda=300/600

lamda =3/6=1/2m

5 0
2 years ago
Stellar nurseries, such as the orion nebula, contain hundreds or more fragmenting and contracting regions, as well as many proto
Natasha_Volkova [10]

The hydrogen fusion process will begin after the protostar reaches a temperature of 10 million degrees kelvin, and it will then turn into a stable star.

<h3>How does a protostar become a stable star?</h3>

The interstellar medium can sometimes be gathered into a large nebula, which is a cloud of gas and dust. A nebula can span a number of light years. These nebulae are where gas and dust can combine to produce stars. Until a star can combine hydrogen into helium, it cannot be considered a star. They are referred to as protostars before then. As gravity starts to gather the gases into a ball, a protostar is created. Accrution is the term for this procedure.

Gravitational energy starts to heat the gasses as gravity draws them into the ball's core, which causes the gasses to radiate radiation. Radiation initially just dissipates into space. However, much of the radiation is retained inside the protostar as it draws in stuff and becomes denser, which causes the protostar to heat up even more quickly.

The hydrogen fusion process will begin after the protostar reaches a temperature of 10 million degrees kelvin, and it will then turn into a star.

Learn more about a protostar here:

brainly.com/question/12534975

#SPJ4

3 0
1 year ago
A typical running track is an oval with 74-mm-diameter half circles at each end. A runner going once around the track covers a d
lisabon 2012 [21]

The centripetal acceleration a is 4.32 \times 10^-4 m/s^2.

<u>Explanation:</u>

The speed is constant and computing the speed from the distance and time for one full lap.

Given, distance = 400 mm = 0.4 m,       Time = 100 s.

Computing the v = 0.4 m / 100 s

                         v = 4 \times 10^-3 m/s.

radius of the circular end r = 37 mm = 0.037 m.

            centripetal acceleration a = v^2 / r

                                                        = (4 \times 10^-3)^2 / 0.037

                                                    a = 4.32 \times 10^-4 m/s^2.

6 0
2 years ago
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