Ask question

Ask question

All categories

3 years ago

14

Two small space probes have been slowed to 10m/s as they approach the moon from the same direction. Probe 1 has a mass of 86kg a

nd probe 2 has a mass of 100kg what is their total momentum

1 answer:

bulgar [2K]3 years ago

4 0

Answer:

B

Explanation:

B

You might be interested in

A 0.50 kg toy is attached to the end of a 1.0 m very light string. The toy is whirled in a horizontal circular path on a frictio

xenn [34]

Answer:

The maximum speed will be 26.475 m/sec

Explanation:

We have given mass of the toy m = 0.50 kg

radius of the light string r = 1 m

Tension on the string T = 350 N

We have to find the maximum speed without breaking the string

For without breaking the string tension must be equal to the centripetal force

So $T=\frac{mv^2}{r}$

So $350=\frac{0.5\times v^2}{1}$

$v^2=700$

v = 26.475 m /sec

So the maximum speed will be 26.475 m/sec

6 0

3 years ago

Which product of nuclear decay has mass but no charge?

Tom [10]

Neutrons have a zero charge but consist of mass.

3 0

3 years ago

What is the Kinetic Energy of a 120 kg object that is moving with a velocity of 15 m/s

gizmo_the_mogwai [7]

Answer:

Kinetic Energy:120 x 15=1800

Explanation:

8 0

3 years ago

Read 2 more answers

Two identical 9.10-g metal spheres (small enough to be treated as particles) are hung from separate 300-mm strings attached to t

Musya8 [376]

Answer:

$n = 1.266\times 10^{12}$

Explanation:

Given data:

mass of sphere is 10 g

Angle between string and vertical axis is $\theta = 13 degree$

thickness of string 300 mm = 0.3 m

$sin\theta =\frac{2}{0.3 m}$

r =0.3 sin 13 = 0.067 m

$Fe = \frac{ kq_1 q-2}{d^2}$

$Fe = \frac{kq^2}{(2r)^2} = mg tan\theta$

$q^2 = mg tan\theta \frac{(2r)^2}{k}$

$= 0.0091 \times 9.8 tan13 \times \frac{(2\times 0.067)^2}{9\times 10^9}$

$q^2 = 4.10\times 10^{-14}$

$q = 2.026 \times 10^{-7} C$

q = ne

$n = \frac{1.6\times 10^{-19}}{2.02\times 10^{-7}}$

$n = 1.266\times 10^{12}$

3 0

3 years ago

two negative charges that are both -3.0 C push each other apart with a force of 19.2 N how far apart are the charges

Hoochie [10]

The electrostatic force between two charges q1 and q2 is given by
$F=k_e \frac{q_1 q_2}{r^2}$
where $k_e =8.99 \cdot 10^9 N m^2 C^{-2}$ is the Coulomb's constant and r is the distance between the two charges.

If we use F=19.2 N and q1=q2=-3.0 C, we can find the value of r, the distance between the two charges by re-arranging the previous formula:
$r= \sqrt{k_e \frac{q_1 q_2}{F} }= \sqrt{ 8.99 \cdot 10^9 N m^2 C^{-2} \frac{(-3.0C)^2}{19.2 N} } =6.49 \cdot 10^4 m=64.9 km$

5 0

3 years ago