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valentina_108 [34]
3 years ago
6

a car has a mass of 1.00 x 10^3 kilograms and it has an acceleration of 4.5 meters/second what is the net force on the car

Physics
2 answers:
love history [14]3 years ago
8 0

Answer:

the answer is 4.5 x10^3 newtons

Explanation:

Reptile [31]3 years ago
6 0
The answer is 4.5 x10^3 newtons.
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. Mass 1 has a momentum of 20 kg*m/s. Mass 2 has a momentum of 50 kg*m/s.
sesenic [268]

Answer:

55kgm/s

Explanation:

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3 years ago
What kind of energy does a rubber band have when it is stretched?
RideAnS [48]
Elastic potential energy. When you stretch a rubber band it has the "potential" to do work, to fly in a given direction. In doing so it changes it's elastic potential energy to kinetic energy.
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3 years ago
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Two loudspeakers are placed next to each other and driven by the same source at 500 Hz. A listener is positioned in front of the
stellarik [79]

To solve this problem we will apply the concepts related to wavelength as the rate of change of the speed of the wave over the frequency. Mathematically this is

\lambda = \frac{v}{f}

Here,

v = Wave velocity

f = Frequency,

Replacing with our values we have that,

\lambda = \frac{340}{500}

\lambda = 0.68m

The distance to move one speaker is half this

\lambda/2 = 0.34m

Therefore the minimum distance will be 0.34m

7 0
3 years ago
Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
Readme [11.4K]

Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

4 0
4 years ago
An equilateral triangle has a height of 3.32 cm. Draw the picture and use this information to determine the length of the sides
Westkost [7]

Answer:

The answer is  a=b=c=3.833 cm

Explanation:

Lets call the variables a=side a b=side b c=side c

We have that the formula of the equilateral triangle for its height is:

1)h=(1/2)*root(3)*a

2) If we resolve the equation we have

 2.1)2h=root(3)*a

 2.2)(2h/root(3))=a

3) After the replacement of each value we have that

a=2*3.32/1.73205

a=3.833 cm

And we know that the equilateral triangle has the same value for each side so a=b=c=3.833 cm

7 0
3 years ago
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