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3 years ago

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A 500N car is pulled up to 20m plant to a flat from 5m above the ground by an effort of 150N parallel to the plank what is the V

R??who can answer!!!!

1 answer:

Stolb23 [73]3 years ago

6 0

Answer:

<h3>4</h3>

Explanation:

VR = Velocity ratio = Distance moved by effort/Distance moved by load

Given:

Distance moved by effort = 20m

Distance moved by Load = 5m

VR = 20/5

VR = 4

Hence the velocity ratio of the car is 4

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A construction worker dropped a brick from a high scaffolding. How fast was? a. How fast was the brick moving after 4.0 s of fal

Zigmanuir [339]

A) How fast was the brick moving after 4s?
Vf=?
Vi=0 (because it was dropped, not thrown)
A= -9.8m/s^2 (gravity)
t= 4s
Use the equation Vf=Vi+A(t)
Vf=0+(-9.8)(4)
Final answer: Vf= -39.2m/s
b) How far did the brick fall after 4s?
D=?
Vi=0
t=4s
A=-9.8m/s2
**You do have the final velocity, but it is best to avoid using numbers that you have calculated yourself.**
Use the equation: d=Vi(t)+0.5(A)(t)^2
d=(0)(4)+0.5(-9.8)(4)^2
d=(-4.9)(16)
d=-78.4m
Therefore, after 4s the brick fell 78.4m

5 0

3 years ago

A 225-kg object and a 525-kg object are separated by 3.80 m. (a) Find the magnitude of the net gravitational force exerted by th

pav-90 [236]

Answer: $F_{net}=3.383\times 10^{-7}\ N$

Explanation:

Given

Mass of first object $m_1=225\ kg$

Mass of second object $m_2=525\ kg$

Distance between them $d=3.8\ m$

$m_3=61\ kg$ object is placed between them

So force exerted by $m_1$ on $m_3$

$F_{13}=\frac{Gm_1m_3}{1.9^2}$

$F_{13}=\frac{6.674\times 10^{-11}(225\times 61)}{1.9^2}$

$F_{13}=2.5374141274×10^{−7}\ N$

Force exerted by $m_2\ on\ m_3$

$F_{23}=\frac{Gm_2m_3}{1.9^2}$

$F_{23}=\frac{6.674\times 10^{-11}(525\times 61)}{1.9^2}$

$F_{23}=5.920632964\times 10^{-7}\ N$

So net force on $m_3$ is

$F_{net}=F_{23}-F_{13}$

$F_{net}=5.920632964\times 10^{-7}-2.5374141274\times 10^{-7}$

$F_{net}=3.383\times 10^{-7}\ N$

i.e. net force is towards $m_2$

(b)For net force to be zero on $m_3$ , suppose

So force exerted by $m_1$ and $m_2$ must be equal

$F_{13}=F_{23}$

$\Rightarrow \frac{Gm_1m_3}{x^2}=\frac{Gm_2m_3}{(3.8-x)^2}$

$\Rightarrow \frac{m_1}{x^2}=\frac{m_2}{(3.8-x)^2}$

$\Rightarrow (\frac{3.8-x}{x})^2=\frac{m_2}{m_1}$

$\Rightarrow \frac{3.8-x}{x}=\sqrt{\frac{525}{225}}$

$\Rightarrow 3.8-x=1.52752x$

$\Rightarrow 3.8=2.52x$

$\Rightarrow x=1.507\ m$

4 0

3 years ago

Why is tungsten used as a filament in a bulb?

Viktor [21]

Answer:

because of tungsten's high melting point

Explanation:

7 0

3 years ago

2. Moving ocean water exerts a force of 375 N on a boat, causing the boat to move a distance of 34.7 m in 8.34 s. What power doe

sasho [114]

1.56 kW , working shown in photo

8 0

3 years ago

3. What does the difference in force depend on?

dangina [55]

Answer:

it depends on the relative masses of the objects.

Explanation:

4 0

3 years ago

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