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2 years ago

8

What tension must a 50.0 cm length of string support in order to whirl an attached 1,000.0 g stone in a circular path at 5.00 m/

s?

1 answer:

galina1969 [7]2 years ago

4 0

Explanation:

Assuming the circular path is horizontal, the sum of forces in the centripetal direction is:

∑F = ma

T = mv²/r

T = (1.0000 kg) (5.00 m/s)² / (0.500 m)

T = 50.0 N

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A shot putter heaves a 7.26kg shot with final velocity of 7.50m/s what is kinetic energy?

ivanzaharov [21]

Answer: KE=.5mv^2

KE=.5(7.62)(7.5)^2

KE=.5(7.62)(56.25)

KE=.5(428.625)

KE=214.3125 J of KE

Explanation:

8 0

2 years ago

A 1.50-m cylindrical rod of diameter 0.500 cm is connected to a power supply that maintains a constant potential difference of 1

nydimaria [60]

Answer:

a) 1.06*10^-5

b) 0.00105 °C^-1

Explanation:

Given that

Length of the cylinder, L = 1.5 m

Radius of the cylinder, r = 0.25 cm

Voltage across the rod, V = 15 V

I• at Temperature T• = 20° C is 18.5 A

I at Temperature T = 90° C is 17.2 A

See attachment for calculations

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2 years ago

A 62kg box is lifted 12 meters off the ground. How much work is done?

Temka [501]

Answer: 7291.2 joules

Explanation:

Work is done when force is applied on an object over a distance.

Thus, Workdone = Force X distance

Since Distance moved by box = 12 metres

mass of box = 62kg

Acceleration due to gravity when box was lifted is represented by g = 9.8m/s^2

Recall that Force = Mass x acceleration due to gravity

i.e Force = 62kg x 9.8m/s^2

= 607.6 Newton

So, Workdone = Force X Distance

Workdone = 607.6 Newton X 12 metres

Workdone = 7291.2 joules

Thus, 7291.2 joules of work was done.

4 0

2 years ago

Answer?????????????????

oksano4ka [1.4K]

the answer of this question is helium

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2 years ago

A person hears a siren as a fire truck approaches and passes by. The frequency varies from 480Hz on approach to 400Hz going away

alekssr [168]

Answer:

31.2 m/s

Explanation:

$f_{app}$ = Frequency of approach = 480 Hz

$f_{aw}$ = Frequency of going away = 400 Hz

$V$ = Speed of sound in air = 343 m/s

$v$ = Speed of truck

Frequency of approach is given as

$f_{app} = \frac{Vf}{V - v}$ eq-1

Frequency of moving awayy is given as

$f_{aw} = \frac{Vf}{V + v}$ eq-2

Dividing eq-1 by eq-2

$\frac{f_{app}}{f_{aw}} = \frac{V + v}{V - v}$

$\frac{480}{400} = \frac{343 + v}{343 - v}$

$v$ = 31.2 m/s

7 0

3 years ago