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Shalnov [3]
2 years ago
9

A child on a 2.4 kg scooter at rest throws a 2.2 kg ball. The ball is given a speed of 3.1 m/s and the child and scooter move in

the opposite directions at 0.45 m/s. Find the child's mass.
Physics
1 answer:
kykrilka [37]2 years ago
3 0

Answer:

The child's mass is 14.133 kg

Explanation:

From the principle of conservation of linear momentum, we have;

(m₁ + m₂) × v₁ + m₃ × v₂ = (m₁ + m₂)  × v₃ - m₃ × v₄

We include the negative sign as the velocities were given as moving in the opposite directions

Since the child and the ball are at rest, we have;

v₁ = 0 m/s and v₂= 0 m/s

Hence;

0 = m₁ × v₃ - m₂ × v₄

(m₁ + m₂)× v₃ = m₃ × v₄

Where:

m₁ = Mass of the child

m₂ = Mass of the scooter = 2.4 kg

v₃ = Final velocity of the child and scooter = 0.45 m/s

m₃ = Mass of the ball = 2.4 kg

v₄ = Final velocity of the ball = 3.1 m/s

Plugging the values gives;

(m₁ + 2.4)× 0.45 = 2.4 × 3.1

(m₁ + 2.4) = 16.533

∴ m₁ + 2.4 = 16.533

m₁ = 16.533 - 2.4 = 14.133 kg

The child's mass = 14.133 kg.

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A puck of mass m = 0.085 kg is moving in a circle on a horizontal frictionless surface. It is held in its path by a massless str
Rina8888 [55]

Answer:

T = 11.93 N

Explanation:

Newton's second law to the puck in the circular path

∑F = m*a Formula (1)

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

m : puck mass  (kg)

a : radial acceleration of the puck (m/s²)

Data:

m = 0.085 kg

L = 0.72 m = R : radium of the circular path (m)

θ=  one revolution = 2Π rad

t= 0.45 s

Angular speed of the puck

ω = θ/t

ω = 1 rev/0.45 s = (2π/0.45) rad/s

ω = 13.96 rad /s

Radial acceleration or centripetal

a = ω²*R

a = (13.96) ²* (0.72)

a = 140.3 m/s²

Magnitude of the tension in the string (T)

We apply the Formula (1)

∑F = m*a

T =  (0.085 kg )*  (140.3 m/s² )

T = 11.93 N

4 0
2 years ago
Consider the waveform expression. y (x, t) = ym sin (0.333x + 5.36 + 585t) The transverse displacement (y) of a wave is given as
Sonja [21]

Explanation:

The waveform expression is given by :

y(x,t)=y_m\ sin(0.333x+5.36+585t)...........(1)

Where

y is the position

t is the time in seconds

The general waveform equation is given by :

y(x,t)=y_m\ sin(kx+\phi+\omega t)..........(2)

Where

k=\dfrac{2\pi}{\lambda}

\omega=2\pi f

On comparing equation (1) and (2) we get :

0.333=\dfrac{2\pi}{\lambda}

\lambda=18.86\ m

585=2\pi f

f = 93.10 Hz

Time period, T=\dfrac{1}{f}

T=\dfrac{1}{0.010}

T = 0.010 s

Phase constant, \phi=5.36\ radian

Hence, this is the required solution.

8 0
2 years ago
You throw a baseball straight up into the air with a speed of 24.5 m/s. How long does it take the baseball to reach its highest
galben [10]
Time=speed/acceleration
Gravitaional Acceleration=9.8 m/s^2
Speed=24.5 m/s
Time=24.5/9.8=2.5 s
3 0
2 years ago
Read 2 more answers
Two wires are used to suspend a sign that weighs 500 N. The two wires make an angle of 100° between each other. If each wire is
vaieri [72.5K]
1) draw a diagram.
2) label diagram. (split the 100 degrees into 50, (which is right down the middle)  to make a right angle triangle.)
3) since its a free body diagram, the forces known must be labelled. (force of gravity). this shows that the straight vertical line of the right angle triangle is Fg (force gravity). label it.
4) use trigonometry. rearrange the equation to solve for what needs to be known.
 
angles known: 50 (split 100 in half to make a right angle triangle), 90 (since its right angle), and 40 (180-90-50 = 40)

sides known: vertical lined up with the 90 degree angle. Fg. --> fg=mg=500N x 9.81m/s^2 = 4905N

use formula: sin or cos 

i used sin. sin(40) = 4905 / ?
- times '?' on both sides. :  sin(40) x '?' = 4905
-divide both sides by sin(40):  '?' = 4905/ sin(40)
--> Solve.

8 0
2 years ago
In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal
SpyIntel [72]

Answer:

A) F_g = 26284.48 N

B) v = 7404.18 m/s

C) E = 19.19 × 10^(10) J

Explanation:

We are given;

Mass of satellite; m = 3500 kg

Mass of the earth; M = 6 x 10²⁴ Kg

Earth circular orbit radius; R = 7.3 x 10⁶ m

A) Formula for the gravitational force is;

F_g = GmM/r²

Where G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Plugging in the relevant values, we have;

F_g = (6.67 × 10^(-11) × 3500 × 6 x 10²⁴)/(7.3 x 10⁶)²

F_g = 26284.48 N

B) From the momentum principle, we have that the gravitational force is equal to the centripetal force.

Thus;

GmM/r² = mv²/r

Making v th subject, we have;

v = √(GM/r)

Plugging in the relevant values;

v = √(6.67 × 10^(-11) × 6 x 10²⁴)/(7.3 x 10⁶))

v = 7404.18 m/s

C) From the energy principle, the minimum amount of work is given by;

E = (GmM/r) - ½mv²

Plugging in the relevant values;

E = [(6.67 × 10^(-11) × 3500 × 6 × 10²⁴)/(7.3 x 10⁶)] - (½ × 3500 × 7404.18)

E = 19.19 × 10^(10) J

5 0
2 years ago
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