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3 years ago

15

A vehicle has an initial velocity of 30 meters per second. 30 seconds later, it is traveling at a velocity of 60 meters per seco

nd. What is the acceleration of the vehicle?

1 answer:

Over [174]3 years ago

3 0

1 m/s^2 (60-30) then divide by 30 secs :)

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Den pushes a desk 400 cm across the floor. He exerts a force of 10 N for 8 s to move the desk. What is his power output? (Power:

White raven [17]

Answer:

5 W

Explanation:

The formula of the power is:

● P = W/t

W is the work and t is the time needed to do it(in seconds)

Let's calculate first the work that the force exerced:

W = Vector F . Vector d

D is the distance ( here 400 cm wich is 4 m)

Make a representation to see how are the vectors F and V.(picture below)

The vector F and d are colinear since Den is pushing the desk on the ground.

● W = 4 × 10 = 40 J

J is Joule

■■■■■■■■■■■■■■■■■■■■■■■■■■

● P = W / t

● P = 40/ 8

● P = 5 W

7 0

3 years ago

Under which of the following conditions is Lactic acid fermentation most likely occur?

adell [148]

Lactic acid is caused by using atp without oxygen being avaliable

5 0

3 years ago

How do you find average velocity

monitta

Answer:

find the sum of the inital and final velocitys and divide by 2 to find the average

4 0

3 years ago

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

So

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

So

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

6 0

3 years ago

Find the total electric charge of 1.7 kg of electrons. me=9.11×10−31kg, e=1.60×10−19C.

Gelneren [198K]

Answer:

$2.99\cdot 10^{11}C$

Explanation:

The mass of one electron is

$m_e = 9.11\cdot 10^{-31}kg$

So the number of electrons contained in M=1.7 kg of mass is

$N=\frac{M}{m_e}=\frac{1.7 kg}{9.11\cdot 10^{-31}kg}=1.87\cdot 10^{30}$

The charge of one electron is

$e=1.60\cdot 10^{-19} C$

So, the total charge of these electrons is equal to the charge of one electron times the number of electrons:

$Q=Ne=(1.87\cdot 10^{30})(1.6\cdot 10^{-19}C)=2.99\cdot 10^{11}C$

8 0

3 years ago