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olga55 [171]
3 years ago
8

Calculate the speed of an object that tavels 62.43m in 38.4s

Physics
1 answer:
TiliK225 [7]3 years ago
8 0
The speed of the object can be calculated using the formula:

Speed = distance/time

The given values are:
distance = 62.43 m
time = 38.4 s

Solution:

speed = 62.43 m / 38.4 s = 1.63 m/s

Therefore, the speed of an object that travels 62.43m in 38.4s is <span> 1.63 m/s.</span>
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What is the speed of sound in air at 50°F (in ft/s)?
gizmo_the_mogwai [7]

Answer:

Speed of air = 1106.38 ft/s

Explanation:

Speed of sound in air with temperature

v_{air}=331.3\sqrt{1+\frac{T}{273.15}} \\

Here speed is in m/s and T is in celcius scale.

T = 50°F

T=(50-32)\times \frac{5}{9}=10^0C \\

Substituting

v_{air}=331.3\sqrt{1+\frac{10}{273.15}}=337.31m/s \\

Now we need to convert m/s in to ft/s.

1 m = 3.28 ft

Substituting

v_{air}=337.31\times 3.28=1106.38ft/s \\

Speed of air = 1106.38 ft/s

6 0
3 years ago
Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?
Hitman42 [59]

The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:

h = \frac{v^{2}}{2g}

<h3>What are perfectly elastic collision?</h3>

Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.

In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.

Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.

Since some of the Kinetic energy is converted to potential energy of the body;

\frac{mv^{2}}{2} = mgh

h = \frac{v^{2}}{2g}

Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.

Learn more about elastic collisions at: brainly.com/question/7694106

7 0
2 years ago
Work output of a large machine in a factory is 89,000 joules, and it’s input is 102,000 joules. Work output of a similar machine
mojhsa [17]
(89000/102000)×100
=87.25%

(92000/104000)×100
=88.46%

efficiency is (output/input)×100
if u get confused which way input and output should go, remember the smaller value is always output and it's above in the fraction, then only it's possible to get a efficiency lower than 100.

7 0
3 years ago
Read 2 more answers
crowbar of 5 metre is used to lift an object of 800 metre if the effort arm is 200cm calculate the force applied​
Klio2033 [76]

Answer:

 F = 5226.6 N

Explanation:

To solve a lever, the rotational equilibrium relation must be used.

We place the reference system on the fulcrum (pivot point) and assume that the positive direction is counterclockwise

         F d₁ = W d₂

where F is the applied force, W is the weight to be lifted, d₁ and d₂ are the distances from the fulcrum.

In this case the length of the lever is L = 5m, t the distance desired by the fulcrum from the weight to be lifted is

d₂ = 200 cm = 2 m

therefore the distance to the applied force is

          d₁ = L -d₂

         d₁ = 5 -2

         d₁= 3m

we clear from the equation

          F = W d₂ / d₁

          W = m g

          F = m g d₂ / d₁

we calculate

      F = 800 9.8 2/3

      F = 5226.6 N

4 0
2 years ago
Basketball player Darrell Griffith is on record as
Gre4nikov [31]

Explanation:

1.

We use the equation

h = \frac{gt^2}{2}, where

h is the height traveled,

g is the acceleration due to gravity and

t is the time taken to reach height h.

We can now calculate t to be

\sqrt{\frac{2*1.2 m}{9.81 m/s^2} }

= 0.495 s

Let v be the initial velocity of the player.

The player deaccelarates from v m/s to 0 m/s in 0.495 s at the rate of 9.81 m/s^2.

v = 9.81 m/s^2 x 0.495 s = 4.85 m/s

2.

The player takes 0.3 s to increase his velocity from 0 m/s to 4.85 m/s. So his average accelaration is

4.85 m/s / 0.3 s = 16.2 m/s^2

5 0
3 years ago
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