Ask question

Ask question

All categories

nikitadnepr [17]

1 year ago

13

A rocket consists of a shuttle and a fuel tank. The rocket flies through the air with a momentum of 180000kgm/s at a velocity of

120m/s. If the fuel tank has a mass of 500kg, what is the mass of the shuttle?

1 answer:

babunello [35]1 year ago

6 0

Answer:

Answer is 1000kg

Hope it help

You might be interested in

An aluminum wire is 5.87 m long and has a diameter of 6.07 mm.

antoniya [11.8K]

Answer: 5.72 x 10-3Ω

Explanation:

Hi, to answer this question, first we have to calculate the cross sectional area of the cable:

Diameter (D)=6.07 mm

Since: 1000mm = 1m

6.07mm/ 1000mm/m = 0.00607 meters

Area of a circle : π (d/2)^2

A = π (0.00607/2)^2= 0.000028937 m2

Resistance formula:

Resistance (R) = P(resistivity) L (length)÷A (cross sectional area )

Replacing with the values given:

R = (2.82x10-8 x 5.87) / 0.000028937

R = 5.72 x 10-3Ω

Feel free to ask for more if needed or if you did not understand something.

7 0

2 years ago

The way your brain interprets the intensity of a sound is the ______.

Naddik [55]

C i would think
it sounds best

8 0

2 years ago

Which change is an example of transforming potential energy to kinetic energy?.

labwork [276]

Answer:

Kinetic energy is energy an object has because of its motion. A ball held in the air, for example, has gravitational potential energy. If released, as the ball moves faster and faster toward the ground, the force of gravity will transfer the potential energy to kinetic energy.

Explanation:

there hope this helps

5 0

2 years ago

If the driver slammed on the brakes, what could happen to the crate?

stich3 [128]

The crate would slide forward

7 0

2 years ago

Read 2 more answers

A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou

zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}$

$E=6.88\times10^{7}\ N/C$

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d). For, r = 7.00 cm

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}$

$E=5.43\times10^{6}\ N/C$

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0

3 years ago