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3 years ago

9

If an object were released in space far away from planted or stars and given an initial momentum, describe what would happen to

that objet

2 answers:

mamaluj [8]3 years ago

7 0

Strange as it may seem, the object would keep moving, in a straight line and at the same speed, until it came near another object. Its momentum and kinetic energy would never change. It might continue like that for a billion years or more.

Have a look at Newton's first law of motion.

-Dominant- [34]3 years ago

4 0

Answer:

The object in motion will stay in motion if there are no other forces affecting it.

Explanation:

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A quarterback claims that he can throw the football a horizontal distance of 167 m. Furthermore, he claims that he can do this b

pychu [463]

Answer:u=42.29 m/s

Explanation:

Given

Horizontal distance=167 m

launch angle $=33.1^{\circ}$

Let u be the initial speed of ball

Range $=\frac{u^2\sin 2\theta }{g}$

$167=\frac{u^2\sin (66.2)}{9.8}$

$u^2=1788.71$

$u=\sqrt{1788.71}$

$u=42.29 m/s$

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3 years ago

Are there any exceptions to the rule that planets rotate with small axis tilts and in the same direction as they orbit the sun?

Alona [7]

<span>Venus, Uranus, and Pluto are exceptions</span>

4 0

3 years ago

What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and

gayaneshka [121]

Force applied on the car due to engine is given as

$F_1 = 300 N$ towards right

Also there is a force on the car towards left due to air drag

$F_2 = 150 N$ towards left

now the net force on the car will be given as

$\vec F_{net} = \vec F_1 + \vec F_2$

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

$F_{net} = F_1 - F_2$

$F_{net} = 300 - 150$

$F_[net} = 150 N$

So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.

5 0

3 years ago

At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco

Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

$A_1*m=M*A_2$

$A_i =Area$

M,m = Counts per second

Our radios are given by

$r_1 = 11cm$

$R_2 = 20cm$

$m = 65cps$

Therefore replacing we have that,

$A_1*m=M*A_2$

$4\pi r_1^2*m = M * 4\pi R_2^2 M$

$r^2*m=MR^2$

$M = \frac{m*r^2}{R^2}$

$M = \frac{65*11^2}{20^2}$

$M = 19.6625cps$

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

7 0

4 years ago

Near the end of a marathon race, the first two runners are separated by a distance of 45.6 m. The front runner has a velocity of

morpeh [17]

Answer:17.08 s

Explanation:

Given

distance between First and second Runner is 45.6 m

speed of first runner $(v_1)$ =3.1 m/s

speed of second runner $(v_2)$ =4.65 m/s

Distance between first runner and finish line is 250 m

Second runner need to run a distance of 250+45.6=295.6 m

Time required by second runner $t=\frac{295.6}{4.65}=63.56 s$

time required by first runner to reach finish line $=\frac{250}{3.1}=80.64 s$

Thus second runner reach the finish line 80.64-63.56=17.08 s earlier

3 0

4 years ago