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Arturiano [62]
3 years ago
6

An object is thrown with a force of 30 Newtons and ends up with an acceleration of 3 m/s ^ 2 due to that throw. What is the mass

of that ball?
Physics
1 answer:
jonny [76]3 years ago
4 0

Answer:

F=30N

a= 3m/s^2

m=?

F=ma

30=m(3)

30/3=m

m=10kg

The mass of the ball is 10kg

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A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f
Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

T=\dfrac{2\pi}{\omega}

{\omega}=\dfrac{2\pi }{0.2}\ rad/s

{\omega}=31.41\ rad/s

We know that

{\omega}^2=m\ K

K=Spring constant

K=\dfrac{\omega^2}{m}

K=\dfrac{31.41^2}{0.12}\ N/m

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

3 0
3 years ago
3.00 textbook rests on a frictionless, horizontal tabletop surface. A cord attached to the book passes over a pulley whose diame
sammy [17]

Answer:

a1 = 3.56 m/s²

Explanation:

We are given;

Mass of book on horizontal surface; m1 = 3 kg

Mass of hanging book; m2 = 4 kg

Diameter of pulley; D = 0.15 m

Radius of pulley; r = D/2 = 0.15/2 = 0.075 m

Change in displacement; Δx = Δy = 1 m

Time; t = 0.75

I've drawn a free body diagram to depict this question.

Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;

ΣF_x = T1 = m1 × a1

a1 is acceleration and can be calculated from Newton's 2nd equation of motion.

s = ut + ½at²

our s is now Δx and a1 is a.

Thus;

Δx = ut + ½a1(t²)

u is initial velocity and equal to zero because the 3 kg book was at rest initially.

Thus, plugging in the relevant values;

1 = 0 + ½a1(0.75²)

Multiply through by 2;

2 = 0.75²a1

a1 = 2/0.75²

a1 = 3.56 m/s²

6 0
2 years ago
Why does gas have the most energy but moves the slowest
Ede4ka [16]

Gases have heavier molecules. Since all gases have the same average kinetic energy at the same temperature, lighter molecules move faster and heavier molecules move slower on average.

4 0
2 years ago
12) Water flows through a horizontal pipe of cross-sectional area 10.0 cm2 at a pressure of 0.250 atm with a flow rate is 1.00 L
masha68 [24]

Answer:

The pressure after passing the valve is 23,8 [Kpa] ( 0,234 atm) and the pressure drop is about 1,53 [Kpa]

Explanation:

We need to use the formula of bernoulli, in the attached image we can see the fluid throw the pipe, we also can calculate the velocity inside the pipe using the flow rate and the cross sectional area.

For this case, we don't use the elevation difference and therefore those terms can be cancelled.

When the area has reduced the velocity of the fluid is increased but there is a drop pressure through the valve.

5 0
3 years ago
Pls help with these two last questions
zhenek [66]

Answer:

F = 50[N], to the left.

v = 10.52 [m/s]

Explanation:

<u>First problem</u>

<u />

In order to solve this problem we must apply Newton's laws, in such a way that we must perform a summation of forces on the horizontal axis. In this way we will analyze each force and the direction of action.

The offensive player is applying a force of 100N to the right, while the defensive player applies a force of 150N to the left. In this way performing the summation of forces we have.

100 - 150 = F

F = - 50 [N]

Note: The negative sign indicates that the resulting force is to the left.

<u>Second problem</u>

<u />

We must remember that the definition of speed is equal to the relationship between distance over time.

x = distance = 100 [m]

t = time = 9.5 [s]

v = x/t

v = 100/9.5

v = 10.52 [m/s]

<u />

5 0
3 years ago
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