What is the power of 10 when 0.00503 is written in scientific notation

The graph of an object's position over time is a horizontal line and y is not equal to 0. What must be true abou

frozen [14]

Answer:D: the velocity is zero

Explanation:

7 0

3 years ago

Which scenario is in excample of a scientific way to think

vlabodo [156]

Can you explain this a bit more I don’t quite understand

4 0

3 years ago

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What

Neko [114]

Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

a

$\alpha = 6261 \ rev/minutes^2$

b

$\theta = 613 \ revolutions$

Explanation:

From the question we are told that

The initial angular speed is $w_i = 1120 \ rev/minutes$

The angular speed after $t = 13.8 s = \frac{13.8}{60 } = 0.23 \ minutes$ is $w_f = 2560 \ rev/minutes$

The time for revolution considered is $t_r = 20 \ s = \frac{20}{60} = 0.333 \ minutes$

Generally the angular acceleration is mathematically represented as

$\alpha = \frac{w_f - w_i }{t}$

=> $\alpha = \frac{2560 - 1120 }{0.23}$

=> $\alpha = 6261 \ rev/minutes^2$

Generally the number of revolution made is $t_r = 20 \ s = \frac{20}{60} = 0.333 \ minutes$ is mathematically represented as

$\theta = \frac{1}{2} * (w_i + w_f)* t$

=> $\theta = \frac{1}{2} * (1120+ 2560 )* 0.333$

=> $\theta = 613 \ revolutions$

5 0

3 years ago

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance

jeka94

Answer:

$1.8\times 105 N/C$

Explanation:

We are given that

$u=2\times 10^7 m/s$

$v=4\times 10^7 m/s$

d=1.9 cm= $\frac{1.9}{100}=0.019 m$

Using 1m=100 cm

We have to find the electric field strength.

$v^2-u^2=2as$

Using the formula

$(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)$

$16\times 10^{14}-4\times 10^{14}=0.038a$

$0.038a=12\times 10^{14}$

$a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2$

$q=1.6\times 10^{-19} C$

Mass of electron,m $=9.1\times 10^{-31} kg$

$E=\frac{ma}{q}$

Substitute the values

$E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}$

$E=1.8\times 105 N/C$

7 0

3 years ago

Doc Brown has calculated his Delorean can accelerate at a rate of 2.52 m/s/s. How

GREYUIT [131]

Answer:

304.89m

Explanation:

Given

acceleration a = 2.52m/s²

final speed v = 39.2m/s

initial speed = 0m/s (car accelerates from rest)

Using the equation of motion below to get the distance of Doc brown from Marty;

v² = u²+2as

substitute the given parameters

39.2² = 0²+2(2.52)s

1536.64 = 0+5.04s

divide both sides by 5.04

1536.64/5.04 = 5.04s/5.04

rearrange the equation

5.04s/5.04 = 1536.64/5.04

s = 304.89m

Hence He and Marty must stand at 304.89m to allow the car to accelerate from rest to a speed of 39.2 m/s?

6 0

3 years ago