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Eva8 [605]
3 years ago
9

How can you determine the amount of work done on an object? (joules, work, power...etc.)

Physics
1 answer:
Novay_Z [31]3 years ago
7 0
Work done is equal to force by distance; so you take the force exerted, in newtons, and multiply that by the direction it's moved (from the starting point in a line, not along the path it's taken.)
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Can you make sound you can see?​
navik [9.2K]
No, sound waves are invisible to our eyes
5 0
3 years ago
"An alpha particle (He2+) with a velocity of 2.6x106m/scrosses a uniform magnetic field at an angle of 37.0°to the field lines a
sergey [27]

Answer:

B = 5.59x10⁹ T

Explanation:

The magnetic force (F), on a the alpha particle with charge (q) that is moving at velocity (v) as the cross product of the velocity and magnetic field (B) is:

F = qvBsin(\theta)

<u>We have:</u>

F = 1.4x10⁻³ N

v = 2.6x10⁶ m/s

θ = 37.0°

q = 2*p = 2*1.6x10⁻¹⁹ C

Hence, the strength of the magnetic field is:

B = \frac{F}{qvsin(\theta)} = \frac{1.4 \cdot 10^{-3}}{1.6 \cdot 10^{-19} C*2.6 \cdot 10^{6}*sin(37)} = 5.59 \cdot 10^{9} T

Therefore, the strength of the magnetic field is 5.59x10⁹ T.

I hope it helps you!

7 0
3 years ago
A 2000 kg car is driving at 5 m/s on wet asphalt, but then makes a turn on some ice and loses control. The driver applies brakes
lesya [120]

Answer:

10,000kgm/s

Explanation:

Since we not told what to look for, we can as well find the momentum of the car.

momentum = mas * velocity

Given

Mass of the car = 2000kg

velocity = 5m/s

Substitute into the formula

Momentum = 2000 * 5

Momentum = 10000kgm/s

Hence the momentum of the car is 10,000kgm/s

5 0
3 years ago
2. A wave on a rope has a wavelength of 2.0 m and a frequency of 2.0 Hz. What is the speed of the
TEA [102]

Answer:

4 m/s or 4 meters per second.

Explanation:

In order to calculate the speed of wave, you multiply the wavelength in meters and the frequency of the Wave in Hertz. 2 times 2 equals 4. The wave speed is always in m/s considering that the wavelength is also in meters.

7 0
4 years ago
The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A
kobusy [5.1K]

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

3 0
4 years ago
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