All categories

3 years ago

A book weighing 12 n is placed on a table. how much support force does a table exert on the book?

Physics

1 answer:

Alenkasestr [34]3 years ago

7 0

It should be 12 N. the force of the book on the table should be the same as the force of the table on the book.

You might be interested in

Your friend from France came to visit you when she was packing she went on weather.com and found that the average temperature in

Marina CMI [18]

Answer:

Because there is not as much cold as it was in France.

Explanation:

The average temperature in France during January ranges from 2.7° to 7.2° celsius which makes it the coldest month of the year. But since she comes to know that average temperature in Annville ranges 31° celsius which implies that the temperature is normal there and therefore, she packs sleeveless tops and shorts. Coats would not be required in a hot weather and hence, she does not pack it.

8 0

3 years ago

Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B i

mel-nik [20]

Answer:

a) αA = 4.35 rad/s²

αB = 1.84 rad/s²

b) t = 3.7 rad/s²

Explanation:

Given:

wA₀ = 240 rpm = 8π rad/s

wA₁ = 8π -αA*t₁

The angle in B is:

$\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2} =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}$

$\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad$

$w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}$

The velocity at the contact point is equal to:

$v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}$

$v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}$

Matching both expressions:

$1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s$

b) The time during which the disks slip is:

$t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s$

a) The angular acceleration of each disk is

$\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)$

$\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)$

6 0

3 years ago

Please help me with my Physical Science! 50 POINTS

DaniilM [7]

Answer:

1, When Jane brakes, the brakes slow the car wheels turning and the road surface exerts a backwards force on the tires, causing the car to decelerate. The pocket book tends to continue on in a straight line (Newton's first law). If she brakes hard enough that the friction between the book and the car seat is insufficient to decelerate the book as fast as the car is decelerating, the book will slide off the seat, and gravity pulls it to the floor

When the diver uses his / her force to depress the springboard, the springboard pushes him back with equal force

3.Newton's Second Law (F=ma)

4. 5 N

5. 19.5 N

65kg * 0.3 m/s^2

6.0.2 N/s

10kg divided by 2N

7.-Walking then pushing the moving forward

-Dribbling

-Basketball is pushed but bounces back

Explanation:

6 0

3 years ago

Read 2 more answers

A box is held at rest by two ropes that form 30° angles with the vertical. The tension T in either rope is 42 N. What is the wei

iragen [17]

<h3><u>Answer;</u></h3>

= 73 N

<h3><u>Explanation</u>;</h3>

Using the formula

2 T cos(30°) = w

Where; T is the tension on each string, while w is the weight of the box given by mg

Therefore;

W = 2Tcos 30°

= 2 × 42 cos 30°

= 84 cos 30°

= 72.74

7 0

3 years ago

Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1360 N. Assume that the play

Alona [7]

Answer:

7.59Ns

Explanation:

Given parameters:

Force = 1360N

Time of contact = 5.85 x 10⁻³s

Unknown:

Impulse = ?

Solution:

The impulse of the ball is given as:

Impulse = Force x time

Impulse = 1360 x 5.85 x 10⁻³ = 7.59Ns

4 0

2 years ago