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2 years ago

An automobile to be transported by ship is raised 8.2m above the dock. If it's GPE is 63,405 J what is the car's mass?

Physics

1 answer:

Pepsi [2]2 years ago

5 0

Answer:

m = 788.2[kg]

Explanation:

The potential energy of a body is defined as the product of mass by gravitational acceleration by height. And it can be calculated by means of the following equation.

$E_{pot}=m*g*h$

where:

Epot = potential energy = 63405 [J]

m = mass [kg]

g = gravity acceleration = 9.81[m/s²]

h = elevation = 8.2[m]

Now replacing:

$63405=m*9.81*8.2\\m=788.2[kg]$

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What velocity must a 1210 kg car have in order to have the same momentum as a pickup truck

Mariulka [41]

Answer: 46.5 m/s

Explanation:

Momentum = mass(m) × velocity (v)

momentum of car = momentum of truck

1210 × v = 2250 × 25

v = 46.5 m/s

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2 years ago

If an object falling freely were somehow equipped with an odometer to measure the distance it travels, then the amount of distan

ioda

Answer:c

Explanation:

Given

object is falling Freely with an odometer

Suppose it falls with zero initial velocity

so distance fallen in time t is given by

$h=ut+\frac{1}{2}gt^2$

here u=0 and t=time taken

$h=\frac{1}{2}gt^2$

for $t=1 s$

$h_1=\frac{1}{2}g$

for $t=2 s$

$h_2=\frac{1}{2}g(2)^2=\frac{4}{2}g=2g$

distance traveled in 2 nd sec $=2g-\frac{1}{2}g=\frac{3}{2}g$

for $t=3 s$

$h_3=\frac{1}{2}g(3)^2=\frac{9}{2}g$

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so we can see that distance traveled in each successive second is increasing

5 0

2 years ago

Three point charges are arranged along the x-axis. Charge q1 = +3.00 uC is at the origin, and charge q2= -5.00 uC is at x= 0.200

artcher [175]

We have all the charges for q1, q2, and q3.
Since k = 8.988x10^2, and N=m^2/c^2

F(1) = F (2on1) + F (3on1)

F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N

Since F1 is 7N,

F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)

Since it wil be going in the negative direction,
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F(3on1) = -10.37N

F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m

So it's located in -.144m

Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.

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