What force is necessary to accelerate a 1050 kg car at a rate of 10 m/s2?

A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass,

Citrus2011 [14]

Answer:

v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

$-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2$

now we know that

$v_0 = 660 m/s$

$M = 5.98 \times 10^{24} kg$

$m = 5 \times 10^9 kg$

$r = 4 \times 10^9 m$

$R = 6.37 \times 10^6 m$

now from above formula

$GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2$

now we have

$2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2$

now plug in all data

$2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2$

$v = 7934.2 m/s$

5 0

3 years ago

A plate drops onto a smooth floor and shatters into three pieces of equal mass.Two of the pieces go off with equal speeds v at r

Firlakuza [10]

Answer:

Speed of the this part is given as

$v_3 = \sqrt2 v$

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

Explanation:

As we know by the momentum conservation of the system

we will have

$P_1 + P_2 + P_3 = P_i$

here we know that

$P_1 = P_2$

the momentum of two parts are equal in magnitude but perpendicular to each other

so we will have

$P_1 + P_2 = \sqrt{P^2 + P^2}$

$P_1 + P_2 = \sqrt2 mv$

now from above equation we have

$P_3 = -(P_1 + P_2)$

$mv_3 = -(\sqrt 2 mv)$

$v_3 = \sqrt2 v$

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

6 0

3 years ago

When a carousel is in motion, the movement of the carousel horse can best be described as?

tatiyna

Supposing the carousel is rotating with constant speed, the movement is uniform angular motion.

5 0

3 years ago

Help me calculate the kinetic energy (just the middle column) ASAP! SHOW WORK! ON PAPER

lukranit [14]

Answer:

Explanation:

I can tell you what the answers for the middle column are, but if you don't know how to solve total energy problems, they won't make any sense to you at all.

First row, KE = 0

Second row, KE = 220500 J

Third row, KE = 183750 J

Fourth row, KE = 205800 J

That's also not paying any attention to significant digits because your velocity only had 1 and that's not enough to do the problem justice. I left all the digits in the answer. Round how your teacher tells you to.

3 0

3 years ago

A steel beam of mass 1975 kg and length 3 m is attached to the wall with a pin that can rotate freely on its right side. A cable

Nuetrik [128]

Answer:

a) 29062.125 N·m

b) 0 N·m

c) $Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}$

d) T = 11186.02 N

Explanation:

We are given

Beam mass = 1975 kg

Beam length = 3 m

Cable angle = 60° above horizontal

a) We have the formula for torque given as follows;

Torque about the pin = Force × Perpendicular distance of force from pin

Where the force = Force due to gravity or weight, we have

Weight = Mass × Acceleration due to gravity = 1975 × 9.81 = 19374.75 N

Point of action of force = Midpoint for a uniform beam = length/2

∴ Point of action of force = 3/2 = 1.5 m

Torque due to gravity = 19374.75 N × 1.5 m = 29062.125 N·m

b) Torque about the pinned end due to the contact forces between the pin and the beam is given by the following relation;

Since the distance from pin to the contact forces between the pin and the beam is 0, the torque which is force multiplied by perpendicular distance is also 0 N·m

c) To find the expression for the tension force, T we find the sum of the moment forces about the pin as follows

Sum of moments about p is given as follows

∑M = 0 gives;

T·sin(θ) × L= M×L/2×g

Therefore torque due to tension is given by the following expression

$Torque, due \ to \ tension =L\cdot Tsin\theta = \frac{M\cdot L\cdot g}{2}$

d) Plugging in the values in the torque due to tension equation, we have;

$3\times Tsin60 = \frac{1975\times 3\times 9.81}{2} = 29062.125$

Therefore, we make the tension force, T the subject of the formula hence

$T= \frac{29062.125}{3 \times sin(60)} = 11186.02 N$

8 0

3 years ago