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egoroff_w [7]
3 years ago
6

What is the acceleration of a 55 kg block of cement when pulled sideways with a net force of 356 N? Answer in units of m/s2.

Physics
1 answer:
puteri [66]3 years ago
5 0
F=ma
a=F/m = 356/55 = .... m/s2
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An observer on the earth sees a spaceship approaching at 0.54c. The ship then launches an exploration vehicle that, according to
AURORKA [14]

Answer:

Explanation:

Expression for relative velocity

= \frac{v_1+v_2}{1+\frac{v_1v_2}{c^2} }

= (.54 + .82 )c/ 1+ \frac{.54 \times.82}{1}

= 1.36 c / 1.4428

= .94 c

β = .94

6 0
3 years ago
A 0.5 kg mass on a spring undergoes simple harmonic motion with a total mechanical energy of 12 J. If the oscillation amplitude
Darya [45]

Answer:

The frequency of the oscillation is 2.45 Hz.

Explanation:

Given;

mass of the spring, m = 0.5 kg

total mechanical energy of the spring, E = 12 J

Determine the spring constant, k as follows;

E = ¹/₂kA²

kA² = 2E

k = (2E) / (A²)

k = (2 x 12) / (0.45²)

k = 118.519 N/m

Determine the angular frequency, ω;

\omega = \sqrt{\frac{k}{m} } \\\\\omega =  \sqrt{\frac{118.519}{0.5} } \\\\\omega = 15.396 \ rad/s

Determine the frequency of the oscillation;

ω = 2πf

f = (ω) / (2π)

f = (15.396) / (2π)

f = 2.45 Hz

Therefore, the frequency of the oscillation is 2.45 Hz.

8 0
2 years ago
Here is a graph of speed vs time. If the object is moving to the east, which BEST describes the speed and velocity of the graph?
Artemon [7]

Answer:

Both speed and velocity are changing.

Explanation:

They are both going up so both are changing

5 0
2 years ago
Two cylindrical rods, one copper and the other iron, are identical in lengths and cross-sectional areas. They are joined, end to
Pie

Answer:

Vc = 2.41 v

Explanation:

voltage (v) = 16 v

find the voltage between the ends of the copper rods .

applying the voltage divider theorem

Vc = V x (\frac{Rc}{Rc + Ri})

where

  • Rc = resistance of copper = \frac{ρl}{a}  (l = length , a = area, ρ = resistivity of copper)
  • Ri = resistance of iron = \frac{ρ₀l}{a}  (l = length , a = area, ρ₀ = resistivity of copper)

Vc =  V x (\frac{\frac{ρl}{a}}{\frac{ρl}{a} + \frac{ρ₀l}{a}})

Vc = V x (\frac{ρ x (\frac{l}{a})}{(ρ + ρ₀) x (\frac{l}{a})})

Vc = V x (\frac{ρ}{ρ + ρ₀})

where

  • ρ = resistivity of copper = 1.72 x 10^{-8} ohm.meter
  • ρ₀ = resistivity of iron = 9.71 x 10^{-8} ohm.meter

Vc = 16 x (\frac{1.72 x 10^{-8}}{1.72 x 10^{-8} + 9.71 x 10^{-8}})

Vc = 2.41 v

5 0
3 years ago
a DJ uses a 110-volt outlet to plug in a strobe light. if the current flowing through the light is 0.050 amps, how much resistan
tigry1 [53]

Ohms Law: V = IR

V is the voltage in volts

I is the current in amps

R is the resistance in Ohms

Rearrange: R = V/I

R = (110)/(0.050)

R = 2200

There are 2200 Ohms of resistance in the circuit.

5 0
3 years ago
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