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4 years ago

The picture below shows an asteroid and a comet.

Physics

1 answer:

Tom [10]4 years ago

6 0

Answer:

01) Asteroids are formed closer to the sun than comets.

Explanation:

Majority of asteroids are present region between Jupiter and Mars while the comets are found away from the orbits of Pluto and Neptune.

You might be interested in

How much weight is generated by an 80 kg person on planet earth? On the moon?

otez555 [7]

It will be 133 on the moon

8 0

3 years ago

A person with a mass of 72 kg is riding a bicycle is accelerating at a rate of 5m/s on a horizontal surface. What is the weight

nikklg [1K]

Answer:

w = 706.32 [N]

Explanation:

The force due to gravitational acceleration can be calculated by means of the product of mass by gravitational acceleration.

w = m*g

where:

w = weight [N] (units of Newtons)

m = mass = 72 [kg]

g = gravity acceleration = 9.81 [m/s²]

Then we have:

$w = 72*9.81\\w = 706.32 [N]$

7 0

3 years ago

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

$F_{D}=\frac{kq^{2}}{DC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

$F_{B}=\frac{kq^{2}}{BC^{2}}$

$F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

$F_{A}=\frac{kq^{2}}{AC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N$

The direction of FA is along A to C.

The net force along +X axis

$F_{x}=F_{A}Cos45-F_{D}$

$F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N$

The net force along +Y axis

$F_{y}=F_{B}-F_{A}Sin45$

$F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N$

The resultant force is given by

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}$

$F = 4.03\times10^{7}N$

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0

3 years ago

Early one October, you go to a pumpkin patch to select your Halloween pumpkin. You lift the 2.9 kg pumpkin to a height of 1.5 m

Katen [24]

Answer:

Work done in lifting to a height of 1.5 m is 42.63 J.

Work done in carrying it to 50 m is 0 J

Explanation:

Given:

Mass of the pumpkin (m) = 2.9 kg

Vertical displacement of the pumpkin (y) = 1.5 m

Horizontal displacement of the pumpkin (x) = 50 m

Acceleration due to gravity (g) = 9.8 m/s²

Work done by a force is given by the formula:

$W=FS\cos\theta$

Where,

$F\to force\ applied\\\\S\to displacement\ caused\\\\\theta\to angle\ between\ F\ and\ S$

Now, the work done is maximum when both force and displacement is in same direction. ( $\theta=0^\circ$ )
Work done is minimum when both force and displacement are in opposite direction. ( $\theta=180^\circ$ )
Work done is zero when force and displacement are perpendicular to each other. ( $\theta=90^\circ$ )

Here, as the pumpkin is raised to a height 'h', there is force applied against gravity in the upward direction. So, force and displacement are in same direction and thus the angle is 0° between the force and displacement vectors.

So, work done is given as:

$W=Force\times Vertical\ displacement$

Force applied is equal to the gravitational force applied by the Earth but in the opposite direction.

So, Force = mg = $2.9\times 9.8 = 28.42\ N$

So, work, $W=Fy=28.42\times 1.5=42.63\ J$

So, work done in lifting the pumpkin is 42.63 J.

Now, when the pumpkin is carried to the check-out stand, the displacement is horizontal while the force applied is still in the upward direction.

So, the force and displacement are perpendicular to each other. Thus, the work done is zero.

4 0

4 years ago

5) Find the initial velocity for a 700 kg car that

Serhud [2]

Answer:

Δv = 12 m/s, but we are not given the direction, so there are really an infinite number of potential solutions.

Maximum initial speed is 40.6 m/s

Minimum initial speed is 16.6 m/s

Explanation:

Assume this is a NET impulse so we can ignore friction.

An impulse results in a change of momentum

The impulse applied was

p = Ft = 1400(6.0) = 8400 N•s

p = mΔv

Δv = 8400 / 700 = 12 m/s

If the impulse was applied in the direction the car was already moving, the initial velocity was

vi = 28.6 - 12 = 16.6 m/s

if the impulse was applied in the direction opposite of the original velocity, the initial velocity was

vi = 28.6 + 12 = 40.6 m/s

Other angles of Net force would result in various initial velocities.

5 0

3 years ago