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3 years ago

Please help and thank you :)

Physics

1 answer:

stellarik [79]3 years ago

3 0

Answer:

20 m/s.

Explanation:

Data obtained from the question include the following:

Initial position (d1) = 20 m

Final position (d2) = 60 m

Initial time (t1) = 1 sec

Final time (t2) = 3 secs

Velocity (v) =?

Next, we shall determine the change in position (Δd). This can be obtained as follow:

Initial position (d1) = 20 m

Final position (d2) = 60 m

Change in position (Δd) =?

Change in position (Δd) = d2 – d1

Change in position (Δd) = 60 – 20

Change in position (Δd) = 40 m

Next, we shall determine the change in time (Δt). This can be obtained as follow:

Initial time (t1) = 1 sec

Final time (t2) = 3 secs

Change in time (Δt) =?

Change in time (Δt) = t2 – t1

Change in time (Δt) = 3 – 1

Change in time (Δt) = 2 secs.

Finally, we shall determine the velocity of the object as follow:

Change in position (Δd) = 40 m

Change in time (Δt) = 2 secs.

Velocity (v) =?

Velocity (v) = Change in position (Δd) /Change in time (Δt)

v = Δd/Δt

v = 40/2

v = 20 m/s

Therefore, the velocity of the object between 1 and 3 secs is 20 m/s

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The tension in a string from which a 4.0 kg object is suspended in an elevator is equal to 44 N. What is the acceleration of the

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Given

Tension is the string $T=44\ N$

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Read 2 more answers

The body of mass 1 kg is kept at rest. A constant force of 6.0 N acting on it the time taken by the body to move through a dista

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Time taken by the body to move through a distance of 12 m is 2 s

Explanation:

using the equation for force

$F= m a$

m= mass= 1 kg

F= force= 6 N

a= acceleration=

6= 1 a

a= 6 m/s²

Now using the kinematic equation,

$d= Vt + \frac{1}{2} at^2$

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V= initial velocity=0 as the object is at rest initially

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3 years ago

When a garden hose with an output diameter of 20 mm is directed straight upward, the stream of water rises to a height of 0.13m

SpyIntel [72]

Answer: h = 0.52m

Explanation:

Using the equation of out flow;

A1 × V1 = A2 ×V2

Where A1 = area of the first nozzle

A2 = area of the second nozzle

V1= velocity of flow out from the first nozzle

V2 = velocity of flow out from 2nd nozzle

But AV= area of nozzle × velocity of water = volume of water per second(m³/s).

Now we can set A×V = Area of nozzle × height of rise.

Henceb A1× h1 = A2 × h2 ( note the time cancel on both sides)

D1 = 20mm= 0.02m; h1 = 0.13m

D2 = 10mm = 0.01m; h2= ?

h2 = π(D1/2)²× h1 /π(D2/2)²

h2 = (0.02/2)² × 0.13/(0.01/2)²

= (0.01)² ×0.13 /(0.005)²

= 1.3 × 10^-5/(5 × 10^-3)²

= 1.3 × 10^-5/25 × 10^-6

= (1.3/25) 10^-5 × 10^6

= 0.052× 10

= 0.52m

7 0

3 years ago