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tekilochka [14]
2 years ago
9

A 9-kg body has three times the kinetic energy of an 4-kg body. Calculate the ratio of the speeds of these bodies.

Physics
1 answer:
Neko [114]2 years ago
3 0

Answer:

 speed 9/speed 4   =   2sqrt3 / 3

Explanation:

KE = 1/2 m v^2

for 9 kg   = 3 * the 4 kg      1/2 (9)va^2   = <u> 3</u>* 1/2 (4) vb^2

va^2 / vb^2 = 3*  1/2 (4) / (1/2 *9)

                      6 / 4.5

va / vb = sqrt (6/4.5) =  2sqrt3 /3

ratio of the 9 speed  to the speed of the 4   is  2 sqrt 3 /3

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A 10cm spring is hung from a clamp stand. With 1N of force it measures 12.2cm long and with 3N of force it measures 16.4cm long.
solmaris [256]

Answer:

0.48 N/cm

Explanation:

The original length is 10cm so with a force of 10cm, the extension will be 12.2-10=2.2 cm when force of 1N is exerted. Similarly, with a force of 3N, the extension is 16.4-10=6.4cm

The slope will be given by change in force/ change in extension and this is equivalent to the stiffness.

Change in force is 3-1=2 N

Change in exgension is 6.4-2.2=4.2 cm

The slope will be 2/4.2=0.4761904761904 N/cm

Rounded off, the slope is 0.48 N/cm and this is the stiffness of the spring.

7 0
3 years ago
A coil with 150 turns and a cross-sectional area of 1.00 m2 experiences a magnetic field whose strength increases by 0.65T in 1.
aleksley [76]

Answer:

54.17volts

Explanation:

Induced emf in a coil placed in a magnetic field can be expressed as E = N¶/t where

N is the number of turns = 150turns

¶ is the magnetic flux = magnetic field strength (B) × area(A)

¶ = BA

B = 0.65T

A = 1.0m²

t is the time =1.8s

Substituting this value in the formula

E = NBA/t

E = 150×0.65×1.0/1.8

E = 54.17Volts

The induced emf in the coil is 54.17Volts

7 0
3 years ago
When you cook a marshmallow on a metal poker tool over an open flame, energy is transferred. Identify the three different ways t
ss7ja [257]

When we cook a marshmallow on a metal poker tool over an open flame, there are three ways in which heat energy is transferred: Conduction, convection, and radiation.

<h3>Heat energy transfer</h3>

Heat transfer is the natural transfer of heat from an object with a higher temperature to an object with a lower temperature. Heat transfer can occur in three ways, namely conduction, convection, and  radiation.

  1. Conduction occurs when heat flows from a place with a high temperature to a place with a lower temperature using a fixed heat-conducting medium. Heat transfer from the open flame to the marshmallows via direct fire contact with the marshmallows is an example of conduction.
  2. Convection is the transfer of heat by means of a stream in which the intermediate substance also moves. If the particles move and cause heat to propagate, convection will occur. The hot air rising from the flames burning the marshmallows is an example of convection.
  3. Radiation is heat transfer without a medium. Radiation can also usually be accompanied by light. The direct transfer of heat from the flame to the marshmallow in the form of waves is an example of radiation.

Learn more about heat transfer here: brainly.com/question/16055406

#SPJ4

3 0
1 year ago
Choose the four major environmental
creativ13 [48]
B that’s the answer your welcome
6 0
3 years ago
Can someone solve this problem and explain to me how you got it​
evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒F\alpha\frac{q1*q2}{r^{2}}

∴F=k\frac{q1*q2}{r^{2}}

where F is the force of attraction or repulsion

k is Coulumb's constant=9*10^{9}Nm^{2}C^{-2}

q1 and q2 are the magnitude of the charges

r is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

F=9*10^{9}*\frac{0.041*0.029}{12^{2}}

F=74312.5N

Question6:

Given: q1=3*10^{-18}C, r=5 m, F=4*10^{-11}N

therefore by Coulumb's law,

4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}

⇒q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C

4 0
3 years ago
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