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svp [43]
3 years ago
14

Lisa is conducting an investigation to determine the coefficient of friction between two surfaces. She uses a 25 kg block. What

is this object's weight, in newtons?
Physics
2 answers:
artcher [175]3 years ago
7 0

Answer:

Weight, W = 245 N

Explanation:

It is given that,

Mass of the block, m = 25 kg

We need to find its weight. From Newton's second law of motion,

F=m\times a

a is the acceleration of the object

On the surface of earth, a=9.8\ m/s^2

F=25\ kg\times 9.8\ m/s^2                    

F = 245 N

So, the weight of the object is 245 N. Hence, this is the required solution.

Paladinen [302]3 years ago
4 0

Answer:

245.25

Explanation:

1 kg is 9.81 newtons

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A 5 kg bowling ball with a velocity of +10 m/s collides with a stationary 2 kg bowling pin. If the ball's final velocity is +8 m
Whitepunk [10]

Answer:

<h2>The pin's final velocity is 5m/s</h2>

Explanation:

Step one:

given data

mass of ball m1=5kg

initial velocity of ball u1=10m/s

mass of pin m2=2kg

initial velocity of pin u2= 0m/s

final velocity of ball v2=8m/s

final velocity of pin v2=?

Step two:

The expression for elastic collision is given as

m1u1+m2u2=m1v1+m2v2

substituting we have

5*10+2*0=5*8+2*v2

50+0=40+2v2

50-40=2v2

10=2v2

divide both sides by 2

v2=10/2

v2=5m/s

The pin's final velocity is 5m/s

3 0
3 years ago
Consider two massless springs connected in series. Spring 1 has a spring constant k1, and spring 2 has a spring constant k2. A c
Andru [333]

Answer:

a. k = (1/k₁ + 1/k₂)⁻¹ b. k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹

Explanation:

Since only one force F acts, the force on spring with spring constant k₁ is F = k₁x₁ where x₁ is its extension

the force on spring with spring constant k₂ is F = k₂x₂ where x₁ is its extension

Let F = kx be the force on the equivalent spring with spring constant k and extension x.

The total extension , x = x₁ + x₂

x = F/k = F/k₁ + F/k₂

1/k = 1/k₁ + 1/k₂

k = (1/k₁ + 1/k₂)⁻¹

B

The force on spring with spring constant k₃ is F = k₃x₃ where x₃ is its extension

Let F = kx be the force on the equivalent spring with spring constant k and extension x.

The total extension , x = x₁ + x₂ + x₃

x = F/k = F/k₁ + F/k₂ + F/k₃

1/k = 1/k₁ + 1/k₂ + 1/k₃

k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹

8 0
2 years ago
Read 2 more answers
A student puts a piece of ice into a beaker of cold water. Which two statements are true. Please help
lesya692 [45]

Answer:

A. Thermal energy will move from the water to the ice.

B. Thermal energy will move from air to the water.

C. Thermal energy will move from the ice to the air.

D. Thermal energy will move from the water to the air.

A and B are correct

Explanation:

Because thermal energy will move from the water to the ice

3 0
2 years ago
How many Americans get sick from consuming contaminated food/water each year
Gnoma [55]

Answer:

48 million people get sick, 128,000 are hospitalized, and 3,000 die

Explanation:

3 0
3 years ago
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A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 34.5 kg.
rodikova [14]

Answer:

the moment of inertia of the merry go round is 38.04 kg.m²

Explanation:

We are given;

Initial angular velocity; ω_1 = 37 rpm

Final angular velocity; ω_2 = 19 rpm

mass of child; m = 15.5 kg

distance from the centre; r = 1.55 m

Now, let the moment of inertia of the merry go round be I.

Using the principle of conservation of angular momentum, we have;

I_1 = I_2

Thus,

Iω_1 = I'ω_2

where I' is the moment of inertia of the merry go round and child which is given as I' = mr²

Thus,

I x 37 = ( I + mr²)19

37I = ( I + (15.5 x 1.55²))19

37I = 19I + 684.7125

37I - 19 I = 684.7125

18I = 684.7125

I = 684.7125/18

I = 38.04 kg.m²

Thus, the moment of inertia of the merry go round is 38.04 kg.m²

7 0
3 years ago
Read 2 more answers
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