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gayaneshka [121]
3 years ago
13

A joule is used to measure a. power. b. energy. c. torque. d. pressure.

Physics
1 answer:
KATRIN_1 [288]3 years ago
5 0
The Joule is the SI unit of energy.
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If an object went from 0 m/s to 6 m/s in 1.7 seconds after a 10 N force was applied to it; what is the object's mass? No links p
Mashcka [7]

The force acting on the object is constant, so the acceleration of the object is also constant. By definition of average acceleration, this acceleration was

<em>a</em> = ∆<em>v</em> / ∆<em>t</em> = (6 m/s - 0) / (1.7 s) ≈ 3.52941 m/s²

By Newton's second law, the magnitude of the force <em>F</em> is proportional to the acceleration <em>a</em> according to

<em>F</em> = <em>m a</em>

where <em>m</em> is the object's mass. Solving for <em>m</em> gives

<em>m</em> = <em>F</em> / <em>a</em> = (10 N) / (3.52941 m/s²) ≈ 2.8 kg

4 0
2 years ago
A box that weighs 5.00×10^2 N is sliding down a ramp at a constant speed. The angle the ramp makes with the horizontal is 25°. W
maxonik [38]

Answer:

0.466 (3 sig. fig.)

Explanation:

Frictional force acting on the box = 5.00×10^2xsin25

Normal force acting on the box = 5.00×10^2xcos25

coefficient of friction = 0.466 (3 sig. fig.)

5 0
3 years ago
A flutist assembles her flute in a room where the speed of sound is 342 m/s. When she plays the note A, it is in perfect tune wi
sertanlavr [38]

Answer:

5.15348 Beats/s

4.55 mm

Explanation:

v_1 = Velocity of sound = 342 m/s

v_2 = Velocity of sound = 346 m/s

f_1 = First frequency = 440 Hz

Frequency is given by

f_2=\frac{v_2}{2L_1}\\\Rightarrow f_2=\frac{346}{2\times 0.38863}\\\Rightarrow f_2=445.15348\ Hz

Beat frequency is given by

|f_1-f_2|=|440-445.15348|=5.15348\ Beats/s

Beat frequency is 5.15348 Hz

Wavelength is given by

\lambda_1=\frac{v_1}{f}\\\Rightarrow \lambda_1=\frac{342}{440}\\\Rightarrow \lambda_1=0.77727\ m

Relation between length of the flute and wavelength is

\lambda_1=2L_1\\\Rightarrow L_1=\frac{\lambda_1}{2}\\\Rightarrow L_1=\frac{0.77727}{2}\\\Rightarrow L_1=0.38863\ m

At v = 346 m/s

\lambda_2=\frac{v_2}{f}\\\Rightarrow \lambda_2=\frac{346}{440}\\\Rightarrow \lambda_1=0.78636\ m

L_2=\frac{\lambda_2}{2}\\\Rightarrow L_2=\frac{0.78636}{2}\\\Rightarrow L_2=0.39318\ m

Difference in length is

\Delta L=L_2-L_1\\\Rightarrow \Delta L=0.39318-0.38863\\\Rightarrow \Delta L=0.00455\ m=4.55\ mm

It extends to 4.55 mm

7 0
3 years ago
Please please help I'm stuck!!!
Korvikt [17]

Answer: iDc im in 3grade

Explanation:sorry

6 0
2 years ago
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Which of the following is true about a planet orbiting a star in uniform circular motion? A. The direction of the velocity vecto
Luda [366]
<span>As it is uniform circular motion therefore speed is constant. Therefore we can rule out option B. Also in circular motion the direction of velocity vector changes therefore velocity can't be constant. Therefore option B is incorrect as well. Also centripetal acceleration is always towards the center so option D is wrong as well. That implies option A is correct.</span>
4 0
3 years ago
Read 2 more answers
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