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3 years ago

Danny is testing out his new golf driver and hits his sister's

Physics

1 answer:

Ivanshal [37]3 years ago

3 0

Answer:

The calculations is shown below on the basis of equation:

work = force × distance

Explanation:

Here the work done is 600 J and distance is certain cm ( assume 10 cm).

As work done is the force applied for certain distance.

Mathematically,

work = force × distance

so,

force = $\frac{work done}{distance}$

or, force = $\frac{600 J}{10 cm}$

or, force= $\frac{600J}{0.1 m}$

or, force = 6000 N

or, force = 6 KN

Hence taking distance as 10 cm, the force will be 6 KN.

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A cat with mass 4.0 kg jumps down to the floor from a bookcase 2.0 m high. What is the cat's change in

eduard

Answer: -7J

Explanation:

To determine the change in potential energy, use the equation ΔPE=mgΔh

PE=mg

, where m is the mass, g= 9.8 m/s2

g= 9.8 m/s

, and Δh is the change in height. Hence, we have that the potential energy PE=4.0 kg×9.8 m/s2×-2.0 m=-78 J

PE=4.0 kg

9.8 m/s

-2.0 m=-78 J

. To verify that this is correct, note that since the cat changes the potential energy to kinetic energy by jumping, the potential energy decreases. Hence, the potential energy should be negative.

3 0

3 years ago

Read 2 more answers

A searchlight is 210 ft from a straight wall. As the beam moves along the wall, the angle between the beam and the perpendicula

Ivenika [448]

Answer:

The length of the beam increasing is 9.64 ft/s.

Explanation:

Given that,

Height = 210 ft

Distance =290 ft

According to figure,

We need to calculate the angle

$\cos\theta=\dfrac{210}{x}$ ....(I)

Put the value of x in the equation

$\cos\theta=\dfrac{210}{290}$

$\cos\theta=\dfrac{21}{29}=0.72$

Now, $\sin\theta=\dfrac{20}{29}$

On differentiate of equation (I)

$-\sin\theta\dfrac{d\theta}{dt}=-\dfrac{-210}{x^2}\dfrac{dx}{dt}$

$\sin\theta=\dfrac{210}{x^2}\dfrac{dx}{dt}$

Put the value in the equation

$\sin\dfrac{20}{29}\times2.0=\dfrac{210}{(290)^2}\dfrac{dx}{dt}$

$\dfrac{dx}{dt}=\sin\dfrac{20}{29}\times2.0\times\dfrac{290^2}{210}$

$\dfrac{dx}{dt}=9.64\ ft/s$

Hence, The length of the beam increasing is 9.64 ft/s.

4 0

3 years ago

a toy airplane is flying at a speed of 8 m/s with an acceleration of 0.9 m/s^2. How fast is it flying after 2 seconds?

Marat540 [252]

Answer:

it will be flying 1.8 m/s

Explanation:

4 0

3 years ago

A garden hose with a diameter of 0.64 in has water flowing in it with a speed of 0.46 m/s and a pressure of 1.9 atmospheres. At

STALIN [3.7K]

Answer:

(a).The speed of the water in the nozzle is 3.014 m/s.

(b). The pressure in the nozzle is 1.86 atm.

Explanation:

Given that,

Nozzle diameter = 0.25 in = 0.00635 m

Hose pipe diameter = 0.64 in = 0.016256 m

Pressure = 1.9 atm =192518 Pa

(a). We need to calculate the speed of the water in the nozzle

Flow Speed at the inlet pipe will be given by using Continuity Equation

$Q_{1}=Q_{2}$

$v_{1}A_{1}=v_{2}A_{2}$

$v_{1}=v_{2}\times(\dfrac{A_{2}}{A_{1}})$

Where, A = area of pipe

$A=\pi\times \dfrac{d^2}{4}$

$v_{1}=v_{2}\times(\dfrac{d_{2}^2}{d_{1}^2})$

Put the value into the formula

$v_{1}=0.46\times\dfrac{(0.016256)^2}{(0.00635)^2}$

$v_{1}=3.014\ m/s$

The speed of the water in the nozzle is 3.014 m/s.

(b). We need to calculate the pressure in the nozzle

Using Bernoulli's Theorem,

$P_{1}+\dfrac{1}{2}\rho\times v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho\times v_{2}^2+\rho gh_{2}$

Where, $h_{1}=h_{2}$

$P_{1}+\dfrac{1}{2}\rho\times v_{1}^2=P_{2}+\dfrac{1}{2}\rho\times v_{2}^2$

$P_{1}=P_{2}+\dfrac{1}{2}\rho(v_{2}^2-v_{1}^2)$

Put the value into the formula

$P_{1}=192518 +\dfrac{1}{2}\times1000\times((0.46)^2-(3.014)^2)$

$P_{1}=188081.702\ Pa$

$P=1.86\ atm$

Hence, (a).The speed of the water in the nozzle is 3.014 m/s.

(b). The pressure in the nozzle is 1.86 atm.

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2 years ago

A batter swings and hits a pitched baseball far over the left field wall. at the moment the baseball contacts the bat, which obj

Lyrx [107]

The baseball bat not the baseball

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3 years ago