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Igoryamba
3 years ago
9

A device used to measure the atmospheric pressure is the barometer. true false

Physics
1 answer:
Angelina_Jolie [31]3 years ago
3 0
True, I just learned this a week ago. Is this for Chemistry?
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How does the force of gravity and the force of earth contribute to africa's poverty?
ivanzaharov [21]

Answer:

The force of gravity is not the same as being on the earth. when your on the earth there no gravitational pull its all up to the air

Explanation:

No explanation

8 0
3 years ago
Recall specific heat of water is 4186 j/kg/C. Find the specific heat of sample.
Paraphin [41]

Answer:

Shown by explanation;

Explanation:

The heat of the sample = mass ×specific heat capacity of the sample × temperature change(∆T)

Assumption;I assume the mass of the samples are : 109g and 192g

∆T= 30.1-21=8.9°c.

The heat of the samples are for 109g are:

0.109 × 4186 × 8.9 =4060.84J

For 0.192g are;

∆T= 67-30.1-=36.9°c

0.192 × 4186×36.9=29656.97J

5 0
3 years ago
A piece of aluminum foil has a known surface density of
bija089 [108]

The cube has 6 equal, square, foil faces. The mass of foil for each face is (380/6) milligrams.

The surface area of each piece is (380)/(6•11) cm^2.

The length of each side of the piece is √(380/6•11) cm

That's about 2.4 cm .

It's a cute little foil cube, just under 1-inch each way.


5 0
3 years ago
A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis (Fig. 11-48).A toy train
masya89 [10]

Answer:

0.166 rad/s

Explanation:

See attachment for calculations

5 0
3 years ago
Read 2 more answers
A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the
Eva8 [605]

Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2= 85.75 J

The speed of the ball after leaving the bat is:

K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}

V=47.92 m/s

Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:

V_f^2-V^2=-2gh

V_f^2-(47.92 m/s)^2=-2*9.81m/s^2*25m

V_f^2=-2*9.81m/s^2*25m+(47.92 m/s)^2

V_f=\sqrt{-2*9.81m/s^2*25m+(47.92 m/s)^2}

V_f=42.5m/s

3 0
3 years ago
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