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2 years ago

If a book has a a mass of 2 kg, how much does the book weigh?

Physics

2 answers:

larisa86 [58]2 years ago

7 0

B. 19 N I got it here

Inessa05 [86]2 years ago

6 0

Answer:

A. 20 N

Explanation:

Weight of the book = mass × acceleration due to gravity

= 2 × 10

= 20 N

You might be interested in

A cannon fires a 0.652 kg shell with initial

laila [671]

Mass have no effect for the projectile motion and u want to know the height "h"
first,
find the vertical and horizontal components of velocity
vertical component of velocity = 12 sin 61
horizontal component of velocity = 12 cos 61
now for the vertical motion ;
S = ut + (1/2) at^2
where
s = h
u = initial vertical component of velocity
t = 0.473 s
a = gravitational deceleration (-g) = -9.8 m/s^2

h=[12×sin 610×0.473]+[−9.8×(0.473)2]

u can simplify this and u will get the answer

h=.5Gt2

H=1.09m

6 0

3 years ago

Read 2 more answers

g A ball thrown straight up into the air is found to be moving at 7.94 m/s after falling 2.72 m below its release point. Find th

kati45 [8]

The ball has height y and velocity v at time t according to

y = v₀ t - 1/2 g t ²

and

v = v₀ - g t

where v₀ is its initial speed and g = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time t such that

-2.72 m = v₀ t - 1/2 g t ²

-7.94 m/s = v₀ - g t

Solve for t in the second equation:

t = (v₀ + 7.94 m/s)/g

Substitute this into the first equation and solve for v₀ :

-2.72 m = v₀ (v₀ + 7.94 m/s) /g - 1/2 g ((v₀ + 7.94 m/s)/g)²

-2.72 m = v₀²/g + (7.94 m/s) v₀/g - 1/2 (v₀ + 7.94 m/s)²/g

2 (-2.72 m) g = 2v₀² + 2 (7.94 m/s) v₀ - (v₀ + 7.94 m/s)²

2 (-2.72 m) (9.80 m/s²) = 2v₀² + (15.9 m/s) v₀ - (v₀² + (15.9 m/s) v₀ + 63.0 m²/s²)

-53.3 m²/s² = v₀² - 63.0 m²/s²

v₀² = 9.73 m²/s²

v₀ = 3.12 m/s

3 0

3 years ago

Work is the transfer of _______ that occurs when a force makes an object move.

Vlad1618 [11]

I think it is energy

4 0

3 years ago

When a voltage difference is applied to a piece of metal wire, a 5.0 mA current flows through it. If this metal wire is now repl

zheka24 [161]

Answer:

I = 21.13 mA ≈ 21 mA

Explanation:

I₁ = 5 mA

L₁ = L₂ = L

V₁ = V₂ = V

ρ₁ = 1.68*10⁻⁸ Ohm-m

ρ₂ = 1.59*10⁻⁸ Ohm-m

D₁ = D

D₂ = 2D

S₁ = 0.25*π*D²

S₂ = 0.25*π*(2*D)² = π*D²

If we apply the equation

R = ρ*L / S

where (using Ohm's Law):

R = V / I

we have

V / I = ρ*L / S

If V and L are the same

V / L = ρ*I / S

then

(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂

S₁ = 0.25*π*D² and

S₂ = 0.25*π*(2*D)² = π*D²

we have

ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)

⇒ I₂ = 4*ρ₁*I₁ / ρ₂

⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m

⇒ I₂ = 21.13 mA

5 0

3 years ago

A push of magnitude P acts on a box of weight W as shown in the figure. The push is directed at an angle θ below the horizontal,

Kipish [7]

Answer:

its on wheels and they are supposed to make it eas

Explanation:

3 0

3 years ago