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Tasya [4]
3 years ago
8

You run away from a plane mirror at 2.30 m/s. At what speed does your image move away from you?

Physics
1 answer:
ivanzaharov [21]3 years ago
5 0

Answer:

4.60m/s

.............

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Find the torque required for the shaft to transmit 40 kW when (a) The shaft speed is 2500 rev/min. (b) The shaft speed is 250 re
Luba_88 [7]

Answer:

(a) 152.85 Nm

(b) 1528.5 Nm

Explanation:

According to the formula of power

P = τ ω

ω = 2 π f

(a) f = 2500 rpm = 2500 / 60 = 41.67 rps

So, 40 x 1000 = τ x 2 x 3.14 x 41.67

τ = 152.85 Nm

(b) f = 250 rpm = 250 / 60 = 4.167 rps

So, 40 x 1000 = τ x 2 x 3.14 x 4.167

τ = 1528.5 Nm

3 0
2 years ago
Speed is determined based on two factors. What are they
a_sh-v [17]

Answer:

distance and time

Explanation: the farther you go and how much time it will take you

7 0
2 years ago
PLSS HELP ITS TIMED <br> WILL MARK BRAINLIEST
dlinn [17]

Answer:

tjfd

Explanation:

tuff

4 0
3 years ago
A camcorder has a power rating of 16 watts. If the output voltage from its battery is 7 volts, what current does it use?
GenaCL600 [577]

Power= current*voltage or P=IV

so 16 watts=I*7 volts

divide on both sides to isolate I so you get

I= 16/7 which is about 2.3 amps

8 0
3 years ago
The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1
Vaselesa [24]

Answer:

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Explanation:

Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:

F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2}) (1)

Where:

F - Explosive force, measured in newtons.

\Delta s - Barrel length, measured in meters.

m - Mass of the shell, measured in kilograms.

v_{o}, v_{f} - Initial and final speeds of the shell, measured in meters per second.

If we know that m = 1250\,kg, v_{o} = 0\,\frac{m}{s}, v_{f} = 750\,\frac{m}{s} and \Delta s = 15\,m, then the explosive force experienced by the shell inside the barrel is:

F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}

F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}

F = 23437500\,N

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

6 0
2 years ago
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