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DanielleElmas [232]
3 years ago
12

Suppose that we replace the aluminum with a mystery metal and repeat the experiment in the video. As in the video, the mass of t

he metal is the same as that of the water. Room temperature is about 20∘C before the start of the experiment. The water heats up to 40∘C, and the mystery metal heats up to 80∘C. Compared to that of water, the heat capacity of our mystery metal is
Physics
1 answer:
katrin2010 [14]3 years ago
7 0

Answer: b) one-third as great.

Explanation:

The options include:

a) three times greater.

b) one-third as great.

c) half as great.

d) two times greater.

e) the same.

Since the water heats up to 40°C and the room temperature is about 20°C before the start of the experiment, heat absorbed will be: (40°C-20°C).= 20°C

Since mystery metal heats up to 80∘C and the room temperature is about 20°C before the start of the experiment, heat absorbed will be: (80°C-20°C).= 60°C.

Therefore, based on the calculation, when compared to that of water, the heat capacity of our mystery metal is (20/60) = 1/3 one-third as great.

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100 points!!! Please help!!!
Greeley [361]

Answer:

41°

Explanation:

Kinetic energy at bottom = potential energy at top

½ mv² = mgh

½ v² = gh

h = v²/(2g)

h = (2.4 m/s)² / (2 × 9.8 m/s²)

h = 0.294 m

The pendulum rises to a height of  above the bottom.  To determine the angle, we need to use trigonometry (see attached diagram).

L − h = L cos θ

cos θ = (L − h) / L

cos θ = (1.2 − 0.294) / 1.2

θ = 41.0°

Rounded to two significant figures, the pendulum makes a maximum angle of 41° with the vertical.

7 0
3 years ago
Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat
Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e \frac{dv}{dt} =10 cm^3/s

Consider r be the radius of the balloon.

The volume of of a sphere is

v=\frac{4}{3} \pi r^3

Differentiate with respect to t

\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}

\Rightarrow 10 =4\pi r^2\frac{dr}{dt}

\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}

The surface of area of the balloon is(S) = 4\pi r^2

S=4\pi r^2

Differentiate with respect to t

\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}

Putting the value of \frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}

\Rightarrow \frac{dS}{dt} =\frac{20}{ r}

Given that r = 5 cm

[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}  =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0
3 years ago
Why are peer reviews important?
riadik2000 [5.3K]
Peer review is important because it is used by scientists to decided which results should be published in a scientific journal
8 0
3 years ago
Read 2 more answers
True or false: Balanced forces can change an object's direction?
slava [35]

The statement is false.  Balanced forces can NOT change the speed OR direction of an object's motion.  (See Newton's #1 law of motion.)

4 0
2 years ago
Read 2 more answers
A 8.34 × 103-kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is
Mnenie [13.5K]

Answer:

21870.3156 N

Explanation:

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 1.6 m/s²

Equation of motion

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-18.2^2}{2\times 162}\\\Rightarrow a=-1.02234\ m/s^2

The acceleration of the craft should be 1.02234 m/s²

F=ma\\\Rightarrow F=8.34\times 10^3\times 1.02234\\\Rightarrow F=8526.3156\ N

Weight of the craft

W=mg\\\Rightarrow W=8.34\times 10^3\times 1.6\\\Rightarrow W=13344\ N

Thrust

F_t=F+W\\\Rightarrow F_t=8526.3156+13344\\\Rightarrow F_t=21870.3156\ N

The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N

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