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Norma-Jean [14]
3 years ago
15

Suppose that a constant force is applied to an object with a mass of 12kg, it’s creates an acceleration of 5m/s^2. The accelerat

ion of another object produced by the same force is 4m/s^2, what is the mass of this object?
Physics
1 answer:
Klio2033 [76]3 years ago
5 0

Answer:

Mass of second object will be 15 kg

Explanation:

We have given mass of first object = 12 kg

Acceleration a=5m/sec^2

According to second law of motion we know that force F = MA

So force F=12\times 5=60N

As the same force is applied to the second object of acceleration a=4m/Sec^2

So force = ma

60=m\times 4

m = 15 kg

So mass of second object will be 15 kg

You might be interested in
A free negative charge released in an electric field will
SashulF [63]

Answer:

Will experience a force due to electric field.

Explanation:

  • When a free negative charge is released in an electric field it experiences a force due to the electric field in a direction opposite to the direction of the magnetic field.

According to Coulomb's law this force is mathematically given as:

F=E.q

and, electric field due to a charge is given as:

E=\frac{1}{4\pi.\epsilon_0}.\frac{q}{r^2}

where:

permittivity of free space\epsilon_0=8.85\times 10^{-12}\ m^{-3}.kg^{-1}.s^4.A^2

q = magnitude of charge

r = radial distance from the charge

5 0
4 years ago
An astronaut standing on a platform on the moon drops a hammer. if the hammer falls 6.0 meters vertically in 2.7 seconds, what i
Amanda [17]
The distance that is traveled by the astronaut given that the motion is free-fall can be calculated through the equation,

    d = Vot + 0.5at²

where d is the distance, Vo is the initial velocity, t is the time, and a is the acceleration. Substituting the known,

   6 = (0 m/s)(2.7 s) + 0.5(a)(2.7 s)²

Determining the value of a,
    a = 1.646 m/s²

ANSWER: 1.646 m/s²
4 0
4 years ago
We can reasonably model a 90-W incandescent lightbulb as a sphere 7.0cm in diameter. Typically, only about 5% of the energy goes
Ronch [10]

Answer:

292.3254055 W/m²

469.26267 V/m

1.56421\times 10^{-6}\ T

Explanation:

P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

r = Radius = \frac{d}{2}=\frac{7}{2}=3.5\ cm

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

c = Speed of light = 3\times 10^8\ m/s

The intensity is given by

I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2

5% of this energy goes to the visible light so the intensity is

I=0.05\times 5846.50811\\\Rightarrow I=292.3254055\ W/m^2

The visible light intensity at the surface of the bulb is 292.3254055 W/m²

Energy density of the wave is

u=\frac{1}{2}\epsilon_0E^2

Energy density is also given by

\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m

The amplitude of the electric field at this surface is 469.26267 V/m

Amplitude of a magnetic field is given by

B=\frac{E}{c}\\\Rightarrow B=\frac{469.26267}{3\times 10^8}\\\Rightarrow B=1.56421\times 10^{-6}\ T

The amplitude of the magnetic field at this surface is 1.56421\times 10^{-6}\ T

7 0
3 years ago
an object is acted upon by a force of 22 Newtons to the right and 13 Newtons to the left what is the magnitude and direction of
frosja888 [35]

22N to the right and 13N to the left

Forces are in opposite directions so you subtract the bigger force from the smaller force

Magnitude of force= 22N -13N

= 9N

net force direction is to the right

8 0
3 years ago
How fast, in meters per second, does an observer need to approach a stationary sound source in order to observe a 7.1 % increase
Nutka1998 [239]

Answer:

v0 = 24.42 m/s (Approx)

Explanation:

Given:

Increase in frequency = 7.1% =

Computation:

Assume n = 100%

n1 = [(v+v0)/(v+v1)]n

[100 + 7.1] =  [(344+v0)/(344+0)]100

107.1 =   [(344+v0)/(344)]100

v0 = 24.42 m/s (Approx)

5 0
3 years ago
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