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jeyben [28]
3 years ago
12

In order to determine an object's speed on speed vs. time graph , the axes MUST be labeled with__

Physics
1 answer:
Mazyrski [523]3 years ago
6 0
<span>b) units and numbers</span>
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A 150 g sample of brass at 100 °C is placed in a Styrofoam cup of water containing 120 mL of water at 10 °C. No heat is lost to
fredd [130]

Answer:

≈19.144°C.

Explanation:

all the details are in the attachment.

Note, that c₁, m₁, t₁ are the parameters of the sample of brass; c₂, m₂ and t₂ are  the parameters of the sample of water.

P.S. change the provided design according Your requirements.

4 0
2 years ago
As light shines from air to water, the index of refraction is 1.02 and the angle of incidence is 38.0 °. What is the light's an
muminat

Answer:

Light's angle of refraction = 37.1° (Approx.)

Explanation:

Given:

Index of refraction = 1.02

Base of refraction = 1

Angle of incidence = 38°

Find:

Light's angle of refraction

Computation:

Using Snell's law;

Sin[Angle of incidence] / Sin[Light's angle of refraction] = Index of refraction / Base of refraction

Sin38 / Light's angle of refraction = 1.02 / 1

Sin[Light's angle of refraction] = Sin 38 / 1.02

Sin[Light's angle of refraction] = [0.6156] / 1.02

Sin[Light's angle of refraction] = 0.6035

Light's angle of refraction = 37.1° (Approx.)

5 0
3 years ago
a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the
ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: F=m*a with this we can get both accelerations; solving for acceleration a=\frac{F}{m}. Now a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2}) anda_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2}). Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, x=\frac{1}{2}*a_{girl}*t^{2} and15.0-x=\frac{1}{2}*a_{sled}*t^{2}, solving for the time we get:t=\sqrt{\frac{2x}{a_{girl} } } and t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } } with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get:\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }. Finally we get:\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} } and replacing the values we have got:\frac{x}{0.14} =\frac{(15.0-x)}{0.73} so 5.33*x=15-x so x=2.37 (meters).

5 0
3 years ago
In an elastic collision, a 580 kg bumper car collides directly from behind with a second, identical bumper car that is traveling
kozerog [31]

Answer:  vl = 2.75 m/s vt = 1.5 m/s

Explanation:

If we assume that no external forces act during the collision, total momentum must be conserved.

If both cars are identical and also the drivers have the same mass, we can write the following:

m (vi1 + vi2) = m (vf1 + vf2) (1)

The sum of the initial speeds must be equal to the sum of the final ones.

If we are told that kinetic energy must be conserved also, simplifying, we can write:

vi1² + vi2² = vf1² + vf2² (2)

The only condition that satisfies  (1) and (2) simultaneously is the one in which both masses exchange speeds, so we can write:

vf1 = vi2 and vf2 = vi1

If we call v1 to the speed of the leading car, and v2 to the trailing one, we can finally put the following:

vf1 = 2.75 m/s  vf2 = 1.5 m/s

8 0
3 years ago
A piano tuner sounds two strings simultaneously. One has been previously tuned to vibrate at 293.0 Hz. The tuner hears 3.0 beats
ololo11 [35]

Answer:

Part a)

f_B = 290 Hz

Part B)

percentage increase is

percentage = 1.38%

Explanation:

Part a)

As we know that the beat frequency is

f_A - f_B = 3

after increasing the tension the beat frequency is decreased and hence the tension in string B will increase

So we have

293 - f_B = 3

f_B = 290 Hz

Part B)

percentage increase in the tension of the string will be given as

f_A - f_B' = 1

f_B' = 292 Hz

now we have

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

so we have

T_1 = C (290)^2

T_2 = C(292)^2

so we have

\frac{\Delta T}{T} = \frac{292^2 - 290^2}{290^2}

percentage increase is

percentage = 1.38

4 0
4 years ago
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