I need to know what order does it go in

. Starting from a stop at a traffic signal, a car speeds up to 20 m/s in 5 seconds. Calculate the acceleration of the car.

Dimas [21]

Answer: 4m/s²

Explanation:

At stop signal car speed iz 0 ; V1=0m/s

After t=5s , car has speed is V2=20m/s

using:

a=(V2-V1)/t

a=(20m/s-0m/s)/5s

a=20m/s/5s

a=4m/s²

4 0

3 years ago

what will be the effect on the acceleration due to gravity of the earth if it is compressed to a the size the moon?

GuDViN [60]

Answer:

it gravitional pull on earth will increased becauste it is compress to a form of moon which is comperatively smaller so the gravitonal pull on per cm of earth will incrased so we can say that there will be change in acceleration due to gravity

7 0

3 years ago

Read 2 more answers

Limestone cave can develop when Limestone rock is weathered. The weathering of the rock leaves an empty space that forms the cav

Ymorist [56]

Answer:

Wind blowing across the surface eroded the limestone.

Explanation:

4 0

3 years ago

Read 2 more answers

A trash can slides on a level floor after being pushed by an unhappy student. It slows and comes to a stop with a constant accel

castortr0y [4]

Answer:

Explanation:

Given:

Acceleration, a = -2.4 m/s^2

Acceleration due to gravity, g = 9.8 m/s^2

Sum of forces = 0

Fr + F = 0

Fr = µsl × m × g

Since, Fn = m × g

m × a = (- µsl × mg)

µsl = -a/g

= 2.4/9.8

= 0.245

5 0

3 years ago

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