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ivann1987 [24]
4 years ago
10

a 10kg cement block horizontally at 6 m/s plows into a bank of sand and comes to a stop in 2.0 m/s. What is the average impact f

orce on the bank?
Physics
1 answer:
Gnesinka [82]4 years ago
7 0
I assume the block plows into the bank of sand with a velocity of 6 m/s and comes to a stop in 2 s.

\bar Ft =  \Delta mv  \\  \\  \bar F =  \frac{\Delta mv }{t}  \\  \\  \bar F \ avarage \ Force \\  m \ momentum \\ v \ velocity \\ t \ time
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You must exert a force of 4.5 N on a book to slide it across a table. If you do 2.7J
hoa [83]

Answer: 0.6m

Explanation:

Given that:

force = 4.5 N

Work done = 2.7J

Distance moved by the book = ?

Since work is done when force is applied on an object over a distance, apply the formula:

work = force x distance

2.7J = 4.5N x distance

Distance = (2.7J / 4.5N)

Distance = 0.6 m

Thus, the book was moved 0.6 metres far

3 0
3 years ago
The position of a full moon is located
marissa [1.9K]

Answer:

opposite the sun. between the Earth and the sun. rising perpendicular to the sun.

Explanation:

4 0
4 years ago
Read 2 more answers
Can you help me??????????????????????????)?
garik1379 [7]

Answer:

Velocity is a change in displacement over change in time and uses the units m/s.

Both are rates of change and can be positive or negative.

Acceleration is a change in velocity over change in time and uses the units m/s².

Explanation:

Velocity is the change in displacement over change in time, this makes it a rate of change. It can be positive or negative because it is a vector quantity. It uses the units m/s because that is a displacement unit over a time unit.

Acceleration is the change in velocity over change in time, this makes it a rate of change. It can be positive or negative because it is also a vector quantity. It uses the units m/s² (m/s/s) because that is a velocity unit over a time unit.

5 0
3 years ago
A veritical brass rod of circular section is loaded by placing a 10 kg wt on top of it .if it's length is 1 m. it's radius of cr
Inga [223]

Answer:

4.37 * 10^-4 J

Explanation:

Energy stored :

mgΔl / 2

m = mass = 10kg ; g = 9.8m/s² ; r = cross sectional Radius = 1cm = 1 * 10-2 m

Δl = mgl / πr²Y

Y = Youngs modulus = Y=3.5 ×10^10 ; l = Length = 1m

Δl = (10 * 9.8 * 1) / π * (1 * 10^-2)²* 3.5 ×10^10

Δl = 98 / 3.5 * π * 10^6

Δl = 0.00000891267

Energy stored :

mgΔl / 2

(10 * 9.8 * 0.00000891267) / 2

= 0.00043672083 J

4.37 * 10^-4 J

3 0
3 years ago
Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular
RUDIKE [14]

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, \omega _i = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

6 0
3 years ago
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