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Montano1993 [528]
3 years ago
11

A wooden block with mass 1.60 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 30.0° (p

oint A). When the spring is released, it projects the block up the incline. At point B, a distance of 6.55 m up the incline from A, the block is moving up the incline at a speed of 7.50 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is \mu_k = 0.50. The mass of the spring is negligible.
Calculate the amount of potential energy that was initially stored in the spring. Take free fall acceleration to be 9.80 m/s².
Physics
1 answer:
andreyandreev [35.5K]3 years ago
7 0

Answer:

The amount of potential energy that was initially stored in the spring is 88.8 J.

Explanation:

Given that,

Mass of block = 1.60 kg

Angle = 30.0°

Distance = 6.55 m

Speed = 7.50 m/s

Coefficient of kinetic friction = 0.50

We need to calculate the amount of potential energy

Using formula of conservation of energy between point A and B

U_{A}+k_{A}+w_{A}=U_{B}+k_{B}

U_{A}+0-fd=mgy+\dfrac{1}{2}mv^2

U_{A}=\mu mg\cos\theta\times d+mg h\sin\theta+\dfrac{1}{2}mv^2

Put the value into the formula

U_{A}=0.50\times1.60\times9.8\cos30\times6.55+1.60\times9.8\times6.55\sin30+\dfrac{1}{2}\times1.60\times(7.50)^2

U_{A}=88.8\ J

Hence, The amount of potential energy that was initially stored in the spring is 88.8 J.

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A very long, straight solenoid with a cross-sectional area of 2.16 cm^2 is wound with 85.7 turns of wire per centimeter. Startin
Zinaida [17]

Answer:

1.74x10⁻⁵ V

Explanation:

n = 85.7 turns/cm => 8570 turns/metre

The field inside the long solenoid is given by B = μ₀ni

B =  4πx10⁻⁷ x 8570 x 0.175t² = 1.884x10⁻³ t²

dB/dt = 3.78x10⁻³ t

 

Cross-sectional Area'A'= 2.16 cm²=> 2.16 x 10^{-4} m²

Now, rate of change of flux linkage  '|Emf|' is given by:

|Emf| = d(NAB)/dt = NA dB/dt

|Emf|  = 5 x 2.16 x 10^{-4} x 3.78x10⁻³ t

|Emf| = 4.0824x10⁻⁶ t

Considering time 't' at which the current = 3.2A , we have

3.2 = 0.175T²

T² = 3.2/0.175

T = 4.28 s

|emf| = 4.0824x10⁻⁶ t  =>  4.0824x10⁻⁶ x4.28

|emf|= 1.74x10⁻⁵ V

Therefore,the magnitude of the emf induced in the secondary winding is 1.74x10⁻⁵ V

7 0
3 years ago
In 2005, the space probe Deep Impact launched a 370 kg projectile into Comet Temple 1. Observing the collision helped scientists
Ilya [14]
Parta a.

Equation: F = G*m1*m2/d^2

Where
F = 32 N
G = 6.67*10^-11 N.m^2/kg^2
m1 = 9.0*10^13kg
m2 =370 kg
d = distance that separate the center of the two objects.

d^2 = G*m1*m2 / F = 6.67*10^-11 N.m^2/kg^2 * 9.0*10^13 kg *370 kg / 32N = 69,409.69 m^2

d = √69,409.69m^2 = 263.5 m

Part B.

The gravitational field of the comet is g = G*m1/d^2

Notice that it does not depend on the mass of other objects.

Notice also that I will use a distance of 5.0 * 10^3 km, because I think that that is the number that you intended to write in the part b. If that is not the number you can put the right number instead because the solution is written step by step.

g = (6.67*10^-11 N*m^2/kg^2)*(9.0*10^13kg)/(5.0*10^3*10^3m)^2 = 2.4*10^-4 N/kg = 2.4*10^-4 m/s^2
3 0
4 years ago
A standard gold bar stored at Fort Knox, Kentucky, is 7.00 inches long, 3.63 inches wide, and 1.75 inches tall. Gold has a densi
Over [174]

Answer:

14.1 kg

Explanation:

Given:

Length=7.00inches

Width=3.63 inches

Height=1.75 inches

density = 19,300 kg/m3.

We can convert the given parameters to metre for unit consistency

But we know 1 inches= 0.0254 metre

✓Then Length l=7.00inches

=7×0.0254 metre=0.1778m

✓Width w =3.63 inches

==3.63 ×0.0254 metre=0.092m

✓Height h =1.75 inches

=1.75 ×0.0254 metre=0.0445 m

But Mass= density × volume

Volume= Length× width×height

Mass= density× Length× width×height

= 19300kg/m³×0.1778×0.0922×0.0445

=14.1 kg

Therefore, the mass of the gold bar is 14.1 kg

6 0
4 years ago
A 70.0-kilogram man is walking at a speed if 2.0 m/s. What is the kinetic energy?
Paraphin [41]

Answer:

140 J

Explanation:

From the the question, the mass of the man =70.0 kg and the speed at which the man is walking =2.0 m/s.

K.E =  \frac{1}{2}m {v}^{2}

where K.E = the kinetic energy, m=mass and v= speed.

By substitution,

K.E =  \frac{1}{2} \times 70 \times  {2}^{2}

\implies \: K.E = \frac{280}{2}

\implies K.E =140

Hence the kinetic energy of the man is 140J

3 0
3 years ago
Two objects are being lifted by a machine. One object has a mass of 2 kg, and is lifted at a speed of 2
kicyunya [14]

Answer:

Kinetic energy = (1/2) (mass) (speed²)

First object: (1/2) (2 kg) (2 m/s)² = 4 joules .

Second object: (1/2) (4 kg) (3 m/s)² = 18 joules .

The second object had more kinetic energy than the first one had.

Explanation:

8 0
3 years ago
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