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3 years ago

5

A 2 kg book is sitting on a table. A 10 N force is pulling it to the right. A 3 N force is pulling to the left. What is the Net

Force acting on the book? A. 2 N down B. 7 N to the right C. 13 N to the right D. 15 N to the right

1 answer:

ella [17]3 years ago

6 0

Here's the 'vector equation' for that situation:

(10N right) + (3N left)

= (10N right) + (-3N right)

= 7N right .

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A Stegosaurus hits' a wall with a force of

Pavel [41]

Answer: a= -1.9 m/s²

Explanation: F = ma and a = F / m = 60 N / 32 kg = 1.875 m/s²

Because of law of action and reaction, wall applies same force to dino

In opposite direction and Dino is decelerating. Thus sign Is negative.

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3 years ago

8. Give the net force for the following:

olya-2409 [2.1K]

Answer:

i think it's is 30N

Explanation:

i got it right

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3 years ago

Which of the following is not accurate when describing solids?

yuradex [85]

The correct answer is: [D]:
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3 years ago

Two blocks of clay, one of mass 100 kg and one of mass 3.00 kg, undergo a completely inelastic collision. Before the collision o

adoni [48]

Answer:

a) v = 0.8 m / s , b) $v_{f}$ = 0.777 m / s , c) ΔK = 0.93 J

Explanation:

This exercise can be solved using the concepts of moment, first let's define the system as formed by the two blocks, so that the forces during the crash have been internal and the moment is conserved.

They give us the mass of block 1 (m1 = 100kg, its kinetic energy (K = 32 J), the mass of block 2 (m2 = 3.00 kg) and that it is at rest (v₀₂ = 0)

Before crash

po = m1 vo1 + m2 vo2

po = m1 vo1

After the crash

$p_{f}$ = (m1 + m2) $v_{f}$

a) The initial speed of the block of m1 = 100 kg, let's use the kinetic energy

K = ½ m v²

v = √2K / m

v = √ (2 32/100)

v = 0.8 m / s

b) The final speed,

p₀ = $p_{f}$

m1 v₀1 = (m1 + m2) $v_{f}$

$v_{f}$ = m1 / (m1 + m2) v₀₁

The initial velocity is calculated in the previous part v₀₁ = v = 0.8 m / s

$v_{f}$ = 100 / (3 + 100) 0.8

$v_{f}$ = 0.777 m / s

c) The change in kinetic energy

Initial K₀ = $K_{f}$

K₀ = 32 J

Final $K_{f}$ = ½ (m1 + m2) $v_{f}$ ²

$K_{f}$ = ½ (3 + 100) 0.777²

$K_{f}$ = 31.07 J

ΔK = $K_{f}$ - K₀

ΔK = 31.07 - 32

ΔK = -0.93 J

As it is a variation it could be given in absolute value

Part D

For this part s has the same initial kinetic energy K = 32 J, but it is block 2 (m2 = 3.00kg) in which it moves

d) we use kinetic energy

v = √ 2K / m2

v = √ (2 32/3)

v = 4.62 m / s

e) the final speed

v₀₂ = v = 4.62 m/s

p₀ = m2 v₀₂

m2 v₀₂ = (m1 + m2) $v_{f}$

$v_{f}$ = m2 / (m1 + m2) v₀₂

$v_{f}$ = 3 / (100 + 3) 4.62

$v_{f}$ = 0.135 m / s

f) variation of kinetic energy

$K_{f}$ = ½ (m1 + m2) $v_{f}$ ²

$K_{f}$ = ½ (3 + 100) 0.135²

$K_{f}$ = 0.9286 J

ΔK = 0.9286-32

ΔK = 31.06 J

4 0

3 years ago

A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi

vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

$V=\frac {C_2V_2-C_1V_1}{C_1+C_2}$

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

$C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}$

Therefore

$V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437$

Charge, Q is given by CV hence for the first capacitor charge will be $Q_1=C_1V$

Here, $Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C$

8 0

3 years ago