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Kruka [31]
2 years ago
13

What is the potential energy when a 100 kg object is raised 4.00 m straight up?

Physics
1 answer:
PtichkaEL [24]2 years ago
8 0

Answer:

100kg x 4 x 10 = 4000J

Explanation:

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During a race, a dragster is 200m from the finish line when something goes wrong and it stops accelerating. It travels at a cons
GarryVolchara [31]

Answer:

135 m

Explanation:

Speed = 45 m/s, time = 3 second

The formula for the distance is given by

Distance = speed x time = 45 x 3 = 135 m

4 0
3 years ago
A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s
Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

  • Time of flight = t = 20 s
  • Angle of the initial velocity of projectile with the horizontal = \theta = 30^\circ

<u>Assume:</u>

  • Initial velocity of the projectile = u
  • R = Range of the projectile during the time of flight
  • H = maximum height of the projectile
  • D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

  • Initial horizontal velocity of the projectile = u\cos \theta
  • Initial horizontal velocity of the projectile = u\sin \theta

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s

Hence, the initial speed  of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

6 0
2 years ago
How much heat is absorbed by a 450g gold block as energy from the sun causes its temperature to change from 27°C to 32°C? (Speci
kotegsom [21]

Answer:

Q = 1057.5 [cal]

Explanation:

In order to solve this problem, we must use the following equation of thermal energy.

Q=m*C_{p}*(T_{final}-T_{initial})

where:

Q = heat energy [cal]

Cp = specific heat = 0.47 [cal/g*°C]

T_final = final temperature = 32 [°C]

T_initial = initial temperature = 27 [°C]

m = mass of the substance = 450 [g]

Now replacing:

Q=450*0.47*(32-27)\\Q=1057.5[cal]

5 0
2 years ago
A large water tank is 3.70 m high and filled to the brim, the top of the tank open to the air. A small pipe with a faucet is att
ipn [44]

h =(3.7 - .58)m  = 3.12m

Now put PE into KE and we have to use the formula:

√2gh (g = gravity and h = height) therefor:

√2 x 9.8 x 3.12

= 7.82m/s

I hope this helps!

7 0
2 years ago
A box of mass m is pushed at an angle 0 by a force Fp along a frictionless surface. The box travels to right with acceleration a
Genrish500 [490]

Answer:A block rests on a horizontal, frictionless surface. A string is attached to the block, and is pulled with a force of 45.0 N at an…

Explanation:

5 0
2 years ago
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