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2 years ago

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When a golf ball is hit, it travels at 41 meters per second. The mass of a golf ball is 0.045kg. Calculate the kinetic energy of

the golf ball.

1 answer:

Fittoniya [83]2 years ago

4 0

Answer:

75.645 J

Explanation:

The kinetic energy is related to the mass and velocity by the formula ...

KE = 1/2mv²

For the given mass of 0.045 kg, and velocity of 41 m/s, the kinetic energy is ...

KE = 1/2(0.045 kg)(41 m/s)² = 75.645 J

__

The unit of energy, joule, is a derived unit equal to 1 kg·m²/s².

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a box is at rest on a ground. An unbalance force acts on the box causes it to start. The box moves 10 meters in 2 seconds and th

nordsb [41]

The box stopped moving because there was a negative acceleration in the x direction caused by friction.
To find the average speed, simply divide the displacement by time: 10 / 2 = 5 m/s

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1 year ago

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Answer:

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Explanation:

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2 years ago

A double nozzle lying in a horizontal x-y plane discharges water into the atmosphere at a rate of 0.5 m3 /s. Assume the water sp

Kisachek [45]

Answer:

The force is $F= 46.25kN$

Explanation:

The diagram for this question is shown on the first uploaded image

At Equilibrium the summation of the of force on the vertical axis is zero

i.e $\sum F_y =0$

=> $F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)$

$v_2$ is the is the speed of water at the nozzle which can be mathematically evaluated as

$v_2 = \frac{R}{A_n}$

substituting $0.5m^3/s$ for R and $\frac{\pi}{4}(12*\frac{1m}{100} )^2$ for $A_n$

$v_2 = \frac{0.5}{\frac{\pi}{4} * (12*\frac{1}{100} )^2 }$

$= 44.23 m/s$

$v_1$ is the is the speed of water at the pipe which can be mathematically evaluated as

$v_1 = \frac{R}{A_p}$

substituting $0.5m^3/s$ for R and $\frac{\pi}{4}(30*\frac{1m}{100} )^2$ for $A_p$

$v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }$

$= 7.07 m/s$

$\rho$ is he density of water with value $\rho =1000 kg /m^3$

Substituting values into the equation above

$F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)$

$= 21.99kN$

At Equilibrium the summation of the of force on the horizontal axis is zero

i.e $\sum F_x =0$

=> $F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)$

Since The speed at both A and B nozzle are the same then $v_2$ remains the same

Substituting values

$F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)$

=> $F_x = 40.69kN$

Hence the force acting on the flange bolts required to hold the nozzle in place is

$F = \sqrt{F_x^2 + F_y^2}$

$= \sqrt{40.69 ^2 + 21.99^2}$

$F= 46.25kN$

6 0

2 years ago

Read 2 more answers

A plane flies 1800 miles in 9 hours, with a tailwind all the way. the return trip on the same route, now with a headwind, tak

Fittoniya [83]

Initially its moving with tail wind so here the speed of wind will support the motion of the plane

so we can say

$V_{plane} + v_{wind} = \frac{distance}{time}$

$V_{plane} + v_{wind} = \frac{1800}{9}$

$V_{plane} + v_{wind} = 200 mph$

now when its moving with head wind we can say that wind is opposite to the motion of the plane

$V_{plane} - v_{wind} = \frac{distance}{time}$

$V_{plane} - v_{wind} = \frac{1800}{12}$

$V_{plane} - v_{wind} = 150mph$

now by using above two equations we can find speed of palne as well as speed of wind

$V_{plane} = 175 mph$

$v_{wind} = 25 mph$

5 0

3 years ago

A block lies on a horizontal frictionless surface. A horizontal force of 100 N is applied to the block giving rise to an acceler

KIM [24]

Answer:

(a) m = 33.3 kg

(b) d = 150 m

(c) vf = 30 m/s

Explanation:

Newton's second law to the block:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

Data

F= 100 N

a= 3.0 m/s²

(a) Calculating of the mass of the block:

We replace dta in the formula (1)

F = m*a

100 = m*3

m = 100 / 3

m = 33.3 kg

Kinematic analysis

Because the block moves with uniformly accelerated movement we apply the following formulas:

d= v₀t+ (1/2)*a*t² Formula (2)

vf= v₀+a*t Formula (3)

Where:

d:displacement in meters (m)

t : time interval in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Data

a= 3.0 m/s²

v₀= 0

t = 10 s

(b) Distance the block will travel if the force is applied for 10 s

We replace dta in the formula (2):

d= v₀t+ (1/2)*a*t²

d = 0+ (1/2)*(3)*(10)²

d =150 m

(c) Calculate the speed of the block after the force has been applied for 10 s

We replace dta in the formula (3):

vf= v₀+a*t

vf= 0+(3*(10)

vf= 30 m/s

4 0

2 years ago