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Nastasia [14]
4 years ago
9

A 3.5 kg book is sitting at rest on a table. what is the value of the normal force being exerted by the table on the book?

Physics
1 answer:
Ira Lisetskai [31]4 years ago
6 0

This is 3.5 * 9.8   (where 9.8 ms-2 is the acceleration due to gravity).

= 34.3 Newtons ( answer).

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Select the correct answer.
Aloiza [94]

Answer:

B

Explanation:

B

7 0
3 years ago
The graph below show how natural processes and human activities affect climate
Kaylis [27]
Is it just me or can anyone else not see a graph? (Because I can’t)
8 0
4 years ago
Read 2 more answers
How much heat energy (in megajoules) is needed to convert 7 kg of ice at -9°C C to water at 0°C?
Molodets [167]

Answer:

hence option A is correct

Explanation:

heat required from -9°C to 0°C ice = mass × specific heat of ice ×change in temperature

heat required from -9°C to 0°C ice = 7×2100×9 =132300 J =0.1323 MJ

( HERE SPECIFIC HEAT OF ICE IS A CONSTANT VALUE OF 2100

J/(kg °C )

heat required from  0°C ice to 0°C water = mass× specific heat of fusion of ice

                                                             = 7×3.36×10^5

                                                              = 2.352 × 10^6 J

                                                              = 2.352 MJ

TOTAL HEAT ENERGY REQUIRED = 0.1323 MJ +2.352 MJ

                                                          = 2.4843 MJ

hence option A is correct

5 0
3 years ago
A loop of wire is placed in a uniform magnetic field. For what orientation of the loop is the magnetic flux a maximum? For what
andreyandreev [35.5K]

Answer:

The flux is calculated as φ=BAcosθ. The flux is thereforemaximum when the magnetic field vector is perpendicular to theplane of the loop. We may also deduce that the flux is zero whenthere is no component of the magnetic field that is perpendicularto the loop.

when angle is zero then flux is maximium because when angle zerocos is maximium

8 0
3 years ago
The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
puteri [66]

Answer:

t=5.3687\ s  is the time taken by the car to accelerate the desired range of the speed from zero at full power.

Explanation:

Given:

Range of speed during which constant power is supplied to the wheels by the car is 0\ mph\ to\ 70\ mph.

  • Initial velocity of the car, v_i=0\ mph
  • final velocity of the car during the test, v_f=32\ mph=14.3052\ m.s^{-1}
  • Time taken to accelerate form zero to 32 mph at full power, t=1.2\ s
  • initial velocity of the car, u_i=0\ mph
  • final desired velocity of the car, u_f=64\ mph=28.6105\ m.s^{-1}

Now the acceleration of the car:

a=\frac{v_f-v_i}{t}

a=\frac{14.3052-0}{1.2}

a=11.921\ m.s^{-1}

Now using the equation of motion:

u_f=u_i+a.t

64=0+11.921\times t

t=5.3687\ s is the time taken by the car to accelerate the desired range of the speed from zero at full power.

8 0
3 years ago
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