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Katena32 [7]
3 years ago
9

A baseball is moving at a speed of 2.2\,\dfrac{\text{m}}{\text{s}}2.2 s m ​ 2, point, 2, space, start fraction, m, divided by, s

, end fraction when it strikes the catcher's glove. The padding of the glove is compressed by 2.4\,\text{mm}2.4mm2, point, 4, space, m, m before the ball comes to a stop. We want to find the average acceleration of the baseball while it is compressing the glove.
Physics
1 answer:
Reika [66]3 years ago
5 0

Answer:

 a = 1,008 10⁻³ m / s²

Explanation:

For this exercise, let's use the kinematic relations of accelerated motion

           v² = v₀² - 2 a x

The negative sign is because the acceleration is opposite to the speed, the final speed is zero

          0 = v₀² - 2 a x

          a = v₀² / 2x

   Let's reduce the magnitudes to the SI system

          x = 2.4mm (1m / 10³mm) = 2.4 10⁻³m

Let's calculate

         a = 2.2²/2 2.4 10⁻³

        a = 1,008 10⁻³ m / s²

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Earth has a magnetic dipole moment of 8.0 × 1022 J/T. (a) What current would have to be produced in a single turn of wire extend
Lelechka [254]

Answer:

Answer is

A. I = 6.3×10^8 A

B. Yes

C. No

Refer below.

Explanation:

Refer to the picture for brief explanation.

7 0
3 years ago
A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles
Dimas [21]
A) The total energy of the system is sum of kinetic energy and elastic potential energy:
E=K+U= \frac{1}{2}mv^2 +  \frac{1}{2}kx^2
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
x=A=4.00 cm = 0.04 m
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
E=U= \frac{1}{2}kA^2 =  \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J

b) When the position of the object is 
x=1.00 cm = 0.01 m
the potential energy of the system is
U= \frac{1}{2}kx^2 =  \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J
and so the kinetic energy is
K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J
since the mass is m=50.0 g=0.05 kg, and the kinetic energy is given by
K= \frac{1}{2}mv^2
we can re-arrange the formula to find the speed of the object:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s

c) The potential energy when the object is at 
x=3.00 cm=0.03 m
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U= \frac{1}{2}kx^2 =  \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J
Therefore the kinetic energy is
K=E-U=0.028 J-0.016 J = 0.012 J

d) We already found the potential energy at point c, and it is given by
U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J
5 0
4 years ago
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3 0
4 years ago
You test a moon buggy on Earth. When the buggy hits a bump, it oscillates up and down on its springs with a period of 4 seconds.
Blizzard [7]

Answer:

Remains same

Explanation:

T = Time period of oscillation

m = mass

k = spring constant

Time period of oscillation is given as

T = 2\pi \sqrt{\frac{m}{k} }

we know that as we move from earth to moon, the value of spring constant "k"  and mass "m" remains unchanged because they do not depend on the acceleration due to gravity.

Time period depends on spring constant inversely and directly on the mass.

hence the time period remains the same.

3 0
4 years ago
HEY CAN ANYONE ANSA DIS!!!
Marizza181 [45]

Answer:

C. 5.35 × 10^-5

Explanation:

5.35 × 0.00001 = 0.0000535

Hope this helped!

8 0
3 years ago
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