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Mrrafil [7]
3 years ago
13

8. An object accelerates 12.0 m/s2 when a force of 6.0 newtons is applied to it. What is the mass of the object? _______________

____________
Physics
1 answer:
Sholpan [36]3 years ago
5 0

Answer:

Mass of object is 0.5kg

Explanation:

Given the following data;

Force = 6N

Acceleration = 12m/s²

Mass =?

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

F = ma

Where;

F represents force.

m represents the mass of an object.

a represents acceleration.

Making mass (m) the subject, we have;

Mass (m) = \frac{F}{a}

Substituting into the equation;

Mass (m) = \frac{6}{12}

Mass, m = 0.5kg.

Therefore, the mass of the object is 0.5kg

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Romashka-Z-Leto [24]

Answer:

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Explanation:

a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)

a= ω^2*r

ω= 300rev/min

convert into rev/s

300/60= 5rev/s

a= 18.75m/s^2

b) use Krot= 1/2 Iω^2

plug in gives

1/2(30*2.25)(25)= 843.75 J

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a lost puppy of 5 kg is 25 m away from his owner of 90 kg. What is the gravitional attraction between the two
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Your local AM radio station broadcasts at a frequency of f = 1100 kHz. The electric-field component of the signal you receive at
Lina20 [59]

Answer:

(a) 272.73 m

(b) 0.338 N/C

Explanation:

frequency, f = 1100 kHz = 1100 x 1000 Hz

E(t) = Eo Sin(2πft)

Eo = 0.62 N/C

(a) Velocity of light, c = 3 x 10^8 m/s

wavelength, λ = c / f = (3 x 10^8) / (1100000) = 272.73 m

Thus, the wavelength is 272.73 m.

(b) at t = 3.1 microsecond = 3.1 x 10^-6 s

E = Eo Sin (2 π ft)

E = 0.62 Sin (2 x 3.14 x 1100 x 10^3 x 3.1 x 10^-6)

E = 0.62 Sin (21.4148)

E = 0.62 x 0.5449 = 0.338 N/C

Thus, the electric field at t = 3.1 microsecond s 0.338 N/C.

6 0
3 years ago
A vertical straight wire 35.0 cmcm in length carries a current. You do not know either the magnitude of the current or whether t
12345 [234]

Answer:

1.714\ \text{A}

Explanation:

F = Magnetic force = 0.018 N

B = Magnetic field = 0.03 T

L = Length of wire = 35 cm

\theta  = Angle between current and magnetic field = 90^{\circ}

Magnetic force is given by

F=IBL\sin\theta\\\Rightarrow I=\dfrac{F}{BL\sin\theta}\\\Rightarrow I=\dfrac{0.018}{0.03\times 35\times 10^{-2}\times \sin90^{\circ}}\\\Rightarrow I=1.714\ \text{A}

The magnitude of the current is 1.714\ \text{A}.

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maxonik [38]
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