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Zigmanuir [339]
3 years ago
6

A book is resting on the table. Forces are exerted _____.

Physics
2 answers:
LekaFEV [45]3 years ago
6 0

Answer:

downward by the book

Explanation:

because of gravity

IRINA_888 [86]3 years ago
4 0

Answer:

up

Explanation:

it feels right

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5 0
3 years ago
Is it possible to put a distance versus time graph to be universal go mine
xz_007 [3.2K]
Yes I'm pretty sure you can
3 0
3 years ago
A dart with mass md is launched toward a block of mass mb that is suspended from a string of length L. The dart is moving horizo
Yuki888 [10]

Answer:

A) Impulse is the same for both the objects

B) The higher is the speed, the greater will be the height.

Explanation:

Part a)

The time of interaction of the two bodies i.e the hanging mass and the stick is same. Thus, force caused by dart on the block = force caused by block on the dart. Hence, impulse is the same for both the objects.  

Part B

The energy will be conserved in the entire reaction process

Hence, Kinetic energy = potential energy

0.5Mv^2 = gh(md+mb)

H is directly proportional to the square of speed.  

Hence, the higher is the speed, the greater will be the height.  

5 0
2 years ago
An astronaut is standing on the surface of a planetary satellite that has a radius of 1.74 × 10^6 m and a mass of 7.35 × 10^22 k
ExtremeBDS [4]

Answer:

2.87 km/s

Explanation:

radius of planet, R = 1.74 x 10^6 m

Mass of planet, M = 7.35 x 10^22 kg

height, h = 2.55 x 10^6 m

G = 6.67 x 106-11 Nm^2/kg^2

Use teh formula for acceleration due to gravity

g=\frac{GM}{R^{2}}

g=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{1.74^{2}\times 10^{12}}

g = 1.62 m/s^2

initial velocity, u = ?, h = 2.55 x 10^6 m , final velocity, v = 0

Use third equation of motion

v^{2}=u^{2}-2gh

0 = v² - 2 x 1.62 x 2.55 x 10^6

v² = 8262000

v = 2874.37 m/s

v = 2.87 km/s

Thus, the initial speed should be 2.87 km/s.

6 0
3 years ago
Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho
77julia77 [94]

Answer:

The time constant is 1.049.

Explanation:

Given that,

Charge q{t}= 0.65 q_{0}

We need to calculate the time constant

Using expression for charging in a RC circuit

q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]

Where, \dfrac{t}{RC} = time constant

Put the value into the formula

0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]

1-e^{-(\dfrac{t}{RC})}=0.65

e^{-(\dfrac{t}{RC})}=0.35

-\dfrac{t}{RC}=ln (0.35)

-\dfrac{t}{RC}=-1.049

\dfrac{t}{RC}=1.049

Hence, The time constant is 1.049.

6 0
3 years ago
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