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Nastasia [14]
3 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]3 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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In comparison with other ocean basins, major sedimentary features such as continental rises and abyssal plains are relatively ra
Maksim231197 [3]

Answer:

Why are continental rises and abyssal plains relatively rare in the Pacific? This is because the extensive system of trenches along the active margins of the Pacific, trap much of the sediments flowing off the continents, preventing them from building the broad, flat abyssal plains typical of the Atlantic ocean basins.

3 0
2 years ago
An object falls from the top of a building that is 25 m high. Air resistance is negligible.
Vlada [557]

The velocity of the object s calculated as 22.1 m/s.

<h3>What is the speed of the object?</h3>

Given that we can write that;

v^2 = u^2 + 2gh

Now u = 0 m/s because the object was dropped from a height

v^2 = 2gh

v = √2 * 9.8 * 25

v = 22.1 m/s

Learn more about velocity:brainly.com/question/18084516

#SPJ1

5 0
2 years ago
Que es la expansión del universo?
anastassius [24]

Answer:

La expansión no es más que el incremento con el tiempo de la distancia entre cualquier par de galaxias lejanas. Se suele utilizar para representar este hecho la analogía de un globo donde hemos pintado una serie de puntos a modo de galaxias.

Explanation:

3 0
2 years ago
A 115 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice.
Hoochie [10]

Answer:

(a) Ff = 0.128 N

(b μk = 0.1135

Explanation:

kinematic analysis

Because the hockey puck  moves with uniformly accelerated movement we apply the following formulas:

vf=v₀+a*t Formula (1)

d= v₀t+ (1/2)*a*t² Formula (2)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s

Calculation of the acceleration of the  hockey puck

We apply the Formula (1)

vf=v₀+a*t      v₀=5.8 m/s ,  vf=0

0=5.8+a*t

-5.8 = a*t

a= -5.8/t   Equation (1)

We replace a= -5.8/t in the Formula (2)

d= v₀*t+ (1/2)*a*t²   ,  d=15.1 m ,  v₀=5.8 m/s

15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²  

15.1= 5.8*t-2.9*t

15.1= 2.9*t

t = 15.1 / 2.9

t= 5.2 s

We replace t= 5.2 s in the equation (1)

a= -5.8/5.2

a= -1.115 m/s²

(a) Calculation of the  frictional force (Ff)

We apply Newton's second law

∑F = m*a    Formula (3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Look at the free body diagram of the  hockey puck in the attached graphic

∑Fx = m*a     m= 115g * 10⁻³ Kg/g = 0.115g    ,  a= -1.12 m/s²

-Ff = 0.115*(-1.115)  We multiply by (-1 ) on both sides of the equation

Ff = 0.128 N

(b) Calculation of the coefficient of friction (μk)

N: Normal Force (N)

W=m*g= 0.115*9.8= 1.127 N : hockey puck  Weight

g: acceleration due to gravity =9.8 m/s²

∑Fy = 0

N-W=0

N = W

N =  1.127 N

μk = Ff/N

μk = 0.128/1.127

μk = 0.1135

8 0
3 years ago
A ball is thrown vertically downward at 10 m/s. what is it’s speed 1s and 2s later
soldi70 [24.7K]

Answer:

20 m/s

30 m/s

Explanation:

Given:

v₀ = -10 m/s

a = -9.8 m/s²

When t = 1 s:

v = v₀ + at

v = (-10 m/s) + (-9.8 m/s²) (1 s)

v = -19.8 m/s

When t = 2 s:

v = v₀ + at

v = (-10 m/s) + (-9.8 m/s²) (2 s)

v = -29.6 m/s

Rounded to one significant figures, the speed of the ball at 1 s and 2 s is 20 m/s and 30 m/s, respectively.

3 0
3 years ago
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