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gayaneshka [121]
3 years ago
10

An oscillating block - spring system has a mechanical energy of 1.10 j, and amplitude of 11.0 cm, and a maximum speed of 1.7 m/s

. find the spring constant.
Physics
1 answer:
ioda3 years ago
3 0
The total mechanical energy of the block-spring system is given by the sum of the potential energy and the kinetic energy of the block:
E=U+K= \frac{1}{2}kx^2 + \frac{1}{2}mv^2
where
k is the spring constant
x is the elongation/compression of the spring
m is the mass of the block
v is the speed of the block

At the point of maximum displacement of the spring, the velocity of the block is zero: v=0, so the kinetic energy is zero and the mechanical energy is just potential energy of the spring:
E= \frac{1}{2}kA^2 (1)
where we used x=A, the amplitude (which is the maximum displacement of the spring).
Since we know 
A = 11.0 cm= 0.11 m
E = 1.10 J
We can re-arrange (1) to find the spring constant:
k= \frac{2E}{A^2} = \frac{2 \cdot 1.10 J}{(0.11 m)^2}=181.8 N/m
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Radius of the loop is 0.18 m or 18 cm

Explanation:

Given :

Current flowing through the wire, I = 45 A

Magnetic field at the center of the wire, B = 1.50 x 10⁻⁴ T

Number of turns in circular wire, N = 1

Consider R be the radius of the circular wire.

The magnetic field at the center of the current carrying circular wire is determine by the relation:

B=\frac{N\mu_{0} I}{2R}

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1.50\times10^{-4} =\frac{4\pi \times10^{-7}\times45 }{2R}

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A toy gun uses a spring to project a 4.5-g soft rubber sphere horizontally. The spring constant is 8.0 N/m, the barrel of the gu
marishachu [46]

Answer:

1.93 m/s

Explanation:

Parameters given:

Mass = 4.5g = 0.0045kg

Spring constant = 8.0 N/m

Length of barrel = 13 cm = 0.013m

Frictional force = 0.035N

Compression = 5.8 cm = 0.058m

First, we find the P. E. stored in the spring:

P. E. = ½*k*x²

P. E. = ½ * 8 * 0.058² = 0.013J

Then, we find the work done by the frictional force while the sphere is leaving the barrel of the gun:

Work = Force * distance

The distance here is the length of the barrel.

Work = 0.035 * 0.13 = 0.0046 J

The kinetic energy of the sphere can now be found:

K. E. = P. E. - Work done

K. E. = 0.013 - 0.0046 = 0.0084J

We can now find the speed using the formula for K. E.:

K. E. = ½*m*v²

0.0084 = ½ * 0.0045 * v²

v² = 0.0084/0.00255 = 3.733

=> v = 1.93 m/s

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sweet [91]

Answer:

The net displacement of the car is 3 km West

Explanation:

Please see the attached drawing to understand the car's trajectory: First in the East direction for 4 km (indicated by the green arrow that starts at the origin (zero), and stops at position 4 on the right (East).

Then from that position, it moves back towards the West going over its initial path, it goes through the origin and continues for 3 more km completing a moving to the West a total of 7 km. This is indicated in the drawing with an orange trace that end in position 3 to the left (West) of zero.

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A car moves in a straight line at 22.0 m/s for 10.0miles, then at 30.0 m/s for another 10.0miles. Calculate the car’s average sp
maw [93]

Answer: 25.38 m/s

Explanation:

We have a straight line where the car travels a total distance D, which is divided into two segments d=10 miles:

D=d+d=2d (1)

Where d=10mi \frac{1609.34 m}{1 mi}=16093.4 m

On the other hand, we know speed is defined as:

S=\frac{d}{t} (2)

Where t is the time, which can be isolated from (2):

t=\frac{d}{S} (3)

Now, for the first segment d=16093.4 m the car has a speed S_{1}=22m/s, using equation (3):

t_{1}=\frac{d}{S_{1}} (4)

t_{1}=\frac{16093.4 m}{22m/s} (5)

t_{1}=731.518 s (6) This is the time it takes to travel the first segment

For the second segment d=16093.4 m the car has a speed S_{1}=30m/s,  hence:

t_{2}=\frac{d}{S_{2}} (7)

t_{2}=\frac{16093.4 m}{30m/s} (8)

t_{2}=536.44 s (9) This is the time it takes to travel the secons segment

Having these values we can calculate the car's average speed S_{ave}:

S_{ave}=\frac{d + d}{t_{1} + t_{2}}=\frac{2d}{t_{1} + t_{2}} (10)

S_{ave}=\frac{2(16093.4 m)}{731.518 s +536.44 s} (11)

Finally:

S_{ave}=25.38 m/s

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