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2 years ago

A 1451 kg car is traveling at 48.0 km/h. How much kinetic energy does it possess? K.E. =

1 answer:

8 0

   Kinetic energy = (1/2) (mass) (speed)²

BUT . . . in order to use this equation just the way it's written,
the speed has to be in meters per second. So we'll have to
make that conversion.

KE = (1/2) · (1,451 kg) · (48 km/hr)² · (1000 m/km)² · (1 hr/3,600 sec)²

   = (725.5) · (48 · 1000 · 1 / 3,600)² (kg) · (km·m·hr / hr·km·sec)²

   = (725.5) · ( 40/3 )² · ( kg·m² / sec²)

   = 128,978 joules (rounded)

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Bulky is working at a UPS loading dock. He boasts he can lift a 154 kg box 4 meters in 30

ivanzaharov [21]

Answer: Power is 200 W

Explanation: Power P = work done / time used.

P = W/t = mgh/t = 154 kg · 9.81 m/s²· 4 m / 30 s = 201 W

8 0

3 years ago

The number of mushrooms needed by a pizza restaurant is given by the

kupik [55]

Answer: 24

Explanation:

Given the following equation:

$g=3n$

Where $g$ is the number of mushrooms in a pizza and $n$ the number of pizzas.

If we know the restaurant will make 8 pizzas ( $n=8$ ), then:

$g=3(8)$

$g=24$ This is the needed number of mushrooms for 8 pizzas

8 0

3 years ago

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 m

kramer

Answer:

The focal length of eyepiece is 3.68 cm.

Explanation:

Given that,

Distance = 19 cm

Focal length = 5.5 mm

Magnification = 200

Object distance = -25 cm

We need to calculate the focal length

Using formula of magnification

$m=\dfrac{d}{f_{o}}+\dfrac{-25}{f_{0}f_{e}}$

Put the value into the formula

$f_{e}=\dfrac{19\times(-25)}{.55(-200-\dfrac{19}{.55})}$

$f_{e}=3.68\ cm$

Hence, The focal length of eyepiece is 3.68 cm.

3 0

3 years ago

A 3-kg rock is thrown upward with a force of 200 N at a location where the local gravitational acceleration is 9.79 m/s2 . Deter

expeople1 [14]

Answer: $56.87m/s^{2}$

Explanation:

If we make an analysis of the net force $F_{net}$ of the rock that was thrown upwards, we will have the following:

$F_{net}=F_{up}-W$ (1)

Where:

$F_{up}=200N$ is the force with which the rock was thrown

$W$ is the weight of the rock

Being the weight the relation between the mass $m=3kg$ of the rock and the acceleration due gravity $g=9.79m/s^{2}$ :

$W=m.g=(3kg)(9.79m/s^{2})$ (2)

$W=29.37 N$ (3)

Substituting (3) in (1):

$F_{net}=200N-29.37 N$ (4)

$F_{net}=170.63 N$ (5) This is the net Force on the rock

On the other hand, we know this force is equal to the multiplication of the mass with the acceleration, according to Newton's 2nd Law:

$F_{net}=m.a$ (6)

Finding the acceleration $a$ :

$a=\frac{F_{net}}{m}$ (7)

$a=\frac{170.63 N}{3kg}$ (8)

Finally:

$a=56.87m/s^{2}$

3 0

2 years ago

A 200kg ball on the end of string is swung in horizontal circle with radius of 0.5m . The ball makes revolution every 2second th

ANEK [815]

Answer: $\dfrac{\pi}{2} ms^{-1}$

Explanation:

Let $T$ be the time required to make one revolution.

Let $r$ be the radius of the circular path.

Let $d$ be the distance travelled by ball in one revolution.

As we know,the distance travelled in one revolution is the circumference of the circle.

So, $d=2\pi r$

Given, $d=0.5m\\T=2sec$

$d=2\times \pi \times 0.5=\pi m$

Speed of an object moving is circular path is define as the ratio of distance travelled in one revolution to the time taken by the object to complete one revolution.

Let $s$ be the speed of the ball.

$s=\frac{d}{T}=\frac{\pi }{2}ms^{-1}$

So,the speed of the ball is $\frac{\pi }{2}ms^{-1}$

5 0

3 years ago