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irga5000 [103]
3 years ago
9

How can density be found

Physics
2 answers:
shusha [124]3 years ago
5 0

Answer:

To find the density of any object, you need to know the Mass (grams) of the object, and its Volume (measured in mL or cm). then divide the mass by the volume in order to get an object's Density.

Explanation:

svet-max [94.6K]3 years ago
5 0
Divide the mass by the volume
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a car accelerates at a constant rate from 15 m/s to 25 m/s while it travels a distance of 125 m. How long does it take to achiev
atroni [7]
T=Vf-Vi/s
25m/s -15m/s/ 125m
10m/s /125m
=0.08s
I hope it’s correct !
6 0
3 years ago
A 0.15 kg baseball is pushed with 100 N force. what will its acceleration be?
sammy [17]

Answer:

The acceleration of the ball is 666.67 m/s²

Explanation:

It is given that,

Mass of the baseball, m = 0.15 kg

Applied force to it, F = 100 N

We need to find the acceleration of the ball. It can be calculated using Newton's second law of motion as :

F = ma

a=\dfrac{F}{m}

a=\dfrac{100\ N}{0.15\ kg}

a=666.67\ m/s^2

So, the acceleration of the ball is 666.67 m/s². Hence, this is the required solution.

8 0
3 years ago
n a distant solar system, a planet of mass 5.0 x 1024 kg orbits a sun of mass 3.0 x 1030 kg at a constant distance of 2.0 x1011
Anna35 [415]

Answer:

F = M2 ω^2 R       centripetal force of sun on planet

ω = (F / (M2 R))^1/2 = 2 pi f = 2 pi / P        where P is the period

P = 2 pi (M2 * R / F)^1/2

F = G M1 M2 / R^2        gravitational force on planet

P = 2 pi {R^3 / (G M1)]^1/2

P = 6.28 [(2.0E11)^3 / (6.67E-11 * 3.0E30)]^1/2

P = 6.28 (8 / 20)^1/2 E7 = 3.9E7 sec

1 yr = 3600 * 24 * 365 = 3.15E7 sec

P = 3.9 / 3.2 = 1.2 years

4 0
3 years ago
A 50.0 kg object is moving at 18.2 m/s when a 200 N force is applied opposite the direction of the objects motion, causing it to
Gekata [30.6K]

Answer:

t = 1.4[s]

Explanation:

To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.

P=m*v\\or\\P=F*t

where:

P = impulse or lineal momentum [kg*m/s]

m = mass = 50 [kg]

v = velocity [m/s]

F = force = 200[N]

t = time = [s]

Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.

(m_{1}*v_{1})-F*t=(m_{1}*v_{2})

where:

m₁ = mass of the object = 50 [kg]

v₁ = velocity of the object before the impulse = 18.2 [m/s]

v₂ = velocity of the object after the impulse = 12.6 [m/s]

(50*18.2)-200*t=50*12.6\\910-200*t=630\\200*t=910-630\\200*t=280\\t=1.4[s]

3 0
3 years ago
Light waves are electromagnetic waves that travel at 3.00 Light waves are electromagnetic waves that travel 108 m/s. The eye is
svlad2 [7]

(a) 5.45 \cdot 10^{14} Hz

The relationship between frequency and wavelength of an electromagnetic wave is given by

c=f \lambda

where

c=3.00 \cdot 10^8 m/s is the speed of light

f is the frequency

\lambda is the wavelength

In this problem, we are considering light with wavelength of

\lambda=5.50 \cdot 10^{-7} m

Substituting into the equation and re-arranging it, we can find the corresponding frequency:

f=\frac{c}{\lambda}=\frac{3.00 \cdot 10^8 m/s}{5.50 \cdot 10^{-7} m}=5.45 \cdot 10^{14} Hz

(b) 1.83\cdot 10^{-15} s

The period of a wave is equal to the reciprocal of the frequency:

T=\frac{1}{f}

And using f=5.45 \cdot 10^{14} Hz as we found in the previous part, we can find the period of this wave:

T=\frac{1}{5.45 \cdot 10^{14} Hz}=1.83\cdot 10^{-15} s

5 0
3 years ago
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