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4 years ago

Identify the action-reaction force pair involved when you catch a ball

2 answers:

8 0

Action: ball hits your hand and is caught. Reaction: kinetic energy is transferred to your body and the ball's velocity is reset to the at rest position.

vladimir1956 [14]4 years ago

7 0

When you catch the ball you act surprised, action- reaction

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Two 20kg spheres are placed with their

Maslowich

Answer:

F = 1.07 x 10⁻⁷ N

Explanation:

The gravitational force of attraction between two objects can be found by the use of Newton's Gravitational Law:

$F = \frac{Gm_{1}m_{2}}{r^2}\\\\$

where,

F = Gravitational Force of attraction = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = m₂ = mass of spheres = 20 kg

r = distance between the objects = 50 cm = 0.5 m

Therefore,

$F = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(20\ kg)(20\ kg)}{(0.5\ m)^2}\\\\$

5 0

3 years ago

Accelerating charges radiate electromagnetic waves. calculate the wavelength of radiation produced by a proton in a cyclotron wi

levacccp [35]

B = magnetic field in the cyclotron = 0.400 T

q = magnitude of charge on a proton = 1.6 x 10⁻¹⁹ C

m = mass of the proton = 1.67 x 10⁻²⁷ kg

f = frequency of revolution of proton in the cyclotron = ?

v = speed of electromagnetic waves = 3 x 10⁸ m/s

λ = wavelength of electromagnetic wave = ?

Frequency of revolution of proton in the cyclotron is given as

f = qB/(2πm)

inserting the values

f = (1.6 x 10⁻¹⁹)(0.400)/(2 (3.14) (1.67 x 10⁻²⁷))

f = 6.1 x 10⁶ Hz

wavelength of electromagnetic wave is given as

λ = v/f

λ = (3 x 10⁸)/(6.1 x 10⁶)

λ = 49.2 m

7 0

3 years ago

Read 2 more answers

At a certain moment, a car travelling at 40mph begins braking for a railroad crossing. Assume thatthe acceleration is constant (

egoroff_w [7]

Answer:

Part a)

$v_f = 17.88 - (0.15) t$

Part b)

$d = 1072.8 m$

Explanation:

As we know that initial speed of the car is

$v = 40 mph$

$v = 40 \times (\frac{1609}{3600})$

$v = 17.88 m/s$

now we have

$v_f = v_i + at$

$0 = 17.88 + a(120)$

$a = -0.15 m/s^2$

Part a)

As we know that acceleration is constant here

so we have

$v_f = v_i + at$

$v_f = 17.88 - (0.15) t$

Part b)

distance moved by the car is given as

$d = \frac{v_F + v_i}{2} t$

now we have

$d = \frac{0 + 17.88}{2}(120)$

$d = 1072.8 m$

3 0

3 years ago

Wanda exerts a constant tension force of 12 N on an essentially massless string to keep a tennis ball (m = 60 g) attached to the

goldfiish [28.3K]

Answer:

r = 0.405m = 40.5cm

Explanation:

In order to calculate the length of the string between Wanda and the ball, you take into account that the tension force is equal to the centripetal force over the ball. So, you can use the following formula:

$F_c=ma_c=m\frac{v^2}{r}$ (1)