All categories

3 years ago

An 6 kg object accelerating from 17 m/s to 10 m/s. What is the change in momentum of the object?

Physics

1 answer:

Oksana_A [137]3 years ago

3 0

Answer:

the change is = 17 × 10 = 170 =28.33

You might be interested in

An electron that has a velocity with x component 1.6 × 106 m/s and y component 2.4 × 106 m/s moves through a uniform magnetic fi

Sergio039 [100]

Answer:

(a) 5.056 x 10^-14 N

(b) 5.056 x 10^-14 N

Explanation:

X component of velocity of electron is 1.6 × 10^6 m/s

Y component of velocity of electron is 2.4 × 10^6 m/s

X component of magnetic field is 0.025 T

Y component of magnetic field is -0.16 T

charge on electron, q = - 1.6 x 10^-19 C

Write the velocity and magnetic field in the vector forms.

$\overrightarrow{v}=1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j}$

$\overrightarrow{B}=0.025\widehat{i}-0.16\widehat{j}$

The force on the charge particle when it is moving in the magnetic field is given by

$\overrightarrow{F}=q\left ( \overrightarrow{v}\times \overrightarrow{B} \right )$

(a) Force on electron is given by

$\overrightarrow{F}=-1.6\times 10^{-19}\left ( 1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j} \right )\times \left ( 0.025\widehat{i}-0.16\widehat{j} \right )$

$\overrightarrow{F}=5.056\times 10^{-14}\widehat{k}$

Magnitude of force is 5.056 x 10^-14 N.

(b) Force on a proton is given by

$\overrightarrow{F}=1.6\times 10^{-19}\left ( 1.6\times 10^{6}\widehat{i}+2.4\times 10^{6}\widehat{j} \right )\times \left ( 0.025\widehat{i}-0.16\widehat{j} \right )$

$\overrightarrow{F}=-5.056\times 10^{-14}\widehat{k}$

Magnitude of force is 5.056 x 10^-14 N.

Thus, the magnitude of force remains same but the direction of force is opposite to each other.

Explanation:

4 0

3 years ago

A satellite travels with a constant speed |v| as it moves around a circle centered on the earth. How much work is done by the gr

7nadin3 [17]

Answer:

W = 0

Explanation:

As the satellite moves in a circle the force is perpendicular to the path, therefore the work that is defined by

W = F. r = f r cos θ

Since the force and the radius are perpendicular, the angle θ = 90º and the cosine 90 = 0, therefore there is no work for the circular motion.

W = 0

3 0

3 years ago

¿Tres ejemplos de rozamiento cinético?

Verizon [17]

Answer:

Un automóvil en movimiento en una carretera.

Una piedra rodando por una colina.

Un hierro que se empuja a través del materia

4 0

3 years ago

In general, what is true of the alpha star in a constellation?

horrorfan [7]

It is generally the brightest compared to the other stars in the constellation.

7 0

4 years ago

An electron moves 4.5 m in the direction of an electric field of strength 325 N/C. What is the change in electrical potential en

Vinvika [58]

Calculate q* E * d
Put q = 1.6 x 10^-19 
E = 325 
d = 4.5

I hope this helps!

6 0

4 years ago