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Maksim231197 [3]

3 years ago

7

John doe gets on the highway in his 1967 Shelby 427 Cobra starting from the dead stop at the bottom of the on ramp of it can be

assumed that John accelerates at a rate of 6.7 m s2 how long does it take for John to reach the speed of 30 m s?

1 answer:

masha68 [24]3 years ago

7 0

Answer:

It will take john 4.477 seconds

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A shell of mass m and speed v explodes into two identical fragments. If the shell was moving horizontally (the positive x direct

-Dominant- [34]

Answer:

The velocity of the other fragment immediately following the explosion is v .

Explanation:

Given :

Mass of original shell , m .

Velocity of shell , + v .

Now , the particle explodes into two half parts , i.e $\dfrac{m}{2}$ .

Since , no eternal force is applied in the particle .

Therefore , its momentum will be conserved .

So , Final momentum = Initial momentum

$mv=\dfrac{mv}{2}+\dfrac{mu}{2}\\\\u=v$

The velocity of the other fragment immediately following the explosion is v .

4 0

2 years ago

If the net force on an object stays the same but the mass of the object is doubled, what will happen to the

Sergio [31]

Answer:

The acceleration of the object decreases I think

Explanation:

5 0

2 years ago

What are the strengths and limitations of the doppler and transit methods? What kind of planets are easiest to detect with each

Arturiano [62]

$\huge\mathfrak\red{✔Answer:-}$

Strength: able to detect planets in a wide range of orbits, as long as orbits aren't face on

Limitations: yield only planet's mass and orbital properties

3 0

3 years ago

When the distance between two stars decreases by one-third, the force between them

pashok25 [27]

Answer:

the force will increase by a factor 2.25

Explanation:

The gravitational force between the two stars is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two stars

r is the distance between the stars

If the distance is decreased by one-third, it means that the new distance is 2/3 of the previous distance

$r'=\frac{2}{3}r$

So the new force will be

$F'=G\frac{m_1 m_2}{(\frac{2}{3}r)^2}=\frac{9}{4} G\frac{m_1 m_2}{r^2}=2.25 F$

So, the force will be 2.25 times the previous value.

4 0

3 years ago

The acceleration due to the earth's gravity, in si units, is 9.8 m/s2. in the absence of air friction, a ball is dropped from re

gogolik [260]

We don't know anything about the amount of distance it travels, but that's okay. The only equation we need here is

velocity(final) = velocity(initial) + acceleration * time
vf = vi + (a * t)

The ball is dropped from rest, so vi = 0 m/s.
We want it so that the ball hits the ground with a final velocity of 60 m/s, so vf = 60 m/s.
We are given the acceleration due to gravity, a = 9.8 m/s^2.
We are solving for the time, t = ?.

Now we just plug in the values.
vf = vi + (a * t)
60 m/s = 0 m/s + (9.8 m/s^2)*(t)

60 = 9.8t

60 / 9.8 = t

t = 6.122 s

Hopefully this is the right answer.

7 0

3 years ago