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dalvyx [7]
3 years ago
14

A ball is dropped from rest.

Physics
1 answer:
nignag [31]3 years ago
6 0

Answer:

How fast is it going? 29.4 Meters per second

How far has it fallen? 44.1 meters

Explanation: Gravitional Acceleration: 9.8 meters per secnd squared!

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Which FBD would represent a car moving right with a motor force of 250 N, and force of friction of 750N, a weight of 8500N and a
babymother [125]

Answer:

Option C

Explanation:

Given that

Motor force is 250 N

Force of friction is 750 N

Weight is 8500 N

And, the normal force is 8500 N

Now based on the above information

Here length of the rector shows the relative magnitude forward force i.e. 250 N i..e lower than the frictional force i.e. backward and weight i.e. 8500 would be equivalent to the normal force

8 0
3 years ago
A train travels 63 kilometers in 2 hours, and then 56 kilometers in 3 hours. What is its average speed? i need the answer and un
kolbaska11 [484]
The average speed is the ratio between the total space and the total time of the motion:
v= \frac{S_{tot}}{t_{tot}}
The total space is 
S_{tot}=63 km + 56 km = 119 km
while the total time is 
t_{tot}=2h+3h = 5h
So, the average velocity is
v= \frac{119 km}{5 h} =23.8 km/h


We can also rewrite it in m/s. The total space is S_{tot}=119000 m, while the time is t_{tot}=5h\cdot 3600 s/h = 18000s, and so
v= \frac{119000m}{18000s}=6.6 m/s
4 0
3 years ago
Please help me with this 29 points
Irina18 [472]

Answer:

)Give the definition of poverty line as defined by the World Bank.

7 0
3 years ago
As a result, the total energy in a ____ system (in other words, a system with no external forces) will remain constant
Alika [10]

Answer:

Isolated or Closed system, both are correct

4 0
3 years ago
After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 m/s. T
motikmotik

Answer:

V' = 0.84 m/s

Explanation:

given,

Linear speed of the ball, v = 2.85 m/s

rise of the ball, h = 0.53 m

Linear speed of the ball, v' = ?

rotation kinetic energy of the ball

KE_r = \dfrac{1}{2}I\omega^2

I of the moment of inertia of the sphere

I = \dfrac{2}{5}MR^2

 v = R ω

using conservation of energy

KE_r = \dfrac{1}{2}( \dfrac{2}{5}MR^2)(\dfrac{V}{R})2

KE_r = \dfrac{1}{5}MV^2

Applying conservation of energy

Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy

\dfrac{1}{2}MV^2 + \dfrac{1}{5}MV^2 = \dfrac{1}{2}MV'^2 + \dfrac{1}{5}MV'^2 + M g h

0.7 V^2 = 0.7 V'^2 + gh

0.7\times 2.85^2 = 0.7\times V'^2 +9.8\times 0.53

V'² = 0.7025

V' = 0.84 m/s

the linear speed of the ball at the top of ramp is equal to 0.84 m/s

6 0
3 years ago
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