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wariber [46]
3 years ago
5

A ball of mass 0.50 kg is fired with velocity 120 m/s into the barrel of a spring gun of mass 1.6 kg initially at rest on a fric

tionless surface. The ball sticks in the barrel at the point of maximum compression of the spring. No energy is lost to friction. What fraction of the ball's initial kinetic energy is stored in the spring?
Physics
1 answer:
Gre4nikov [31]3 years ago
4 0

Answer:

The fraction of the ball's kinetic energy stored in the spring is 3.6KJoules

Explanation:

Given

Mass of ball =0.5kg

Velocity of ball =120m/s

The kinetic energy stored by the ball is expressed as

K.E=1/2(m*v²)

Substituting our data into the expression we have

K. E=(0.5*120²)/2

K. E=7200/2

K.E=3,600 JOULE

K.E=3.6KiloJoules

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A grandfather clock is controlled by a swinging brass pendulum that is 1.2 m long at a temperature of 27°C. (a) What is the leng
stealth61 [152]

Answer:

L2 = 1.1994 m

the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m

Explanation:

Given;

Initial length L1 = 1.2m

Initial temperature T1 = 27°C

Final temperature T2 = 0.0°C

Linear expansion coefficient of brass x = 1.9 × 10^-5 /°C

The change i length ∆L;

∆L = L2 - L1

L2 = L1 + ∆L ...........1

∆L = xL1(∆T)

∆L = xL1(T2 - T1) ......2

Substituting the given values into equation 2;

∆L = 1.9 × 10^-5 /°C × 1.2m × (0 - 27)

∆L = 1.9 × 10^-5 /°C × 1.2m × (- 27)

∆L = -6.156 × 10^-4 m

From equation 1;

L2 = L1 + ∆L

Substituting the values;

L2 = 1.2 m + (- 6.156 × 10^-4 m)

L2 = 1.2 m - 6.156 × 10^-4 m

L2 = 1.1993844 m

L2 = 1.1994 m

the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m

3 0
3 years ago
I need help like right now!!!!! please
Mariana [72]

1. All of the above

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3. Nonpolluting & Can be developed anywhere

4. Protons and Neutrons

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5 0
3 years ago
11. A disk 8.00 cm in radiu: rotates at a constant rate of 1200 revinin about its central axis Determine (a) its angular speed i
marshall27 [118]

Answer:

A disk 8.00 cm in radiu: rotates at a constant rate of 1200 revinin about its central axis Determine (a) its angular speed in zadians per second, (b) the tangential speed at a point 3.00 cm from its center, (c) the radial aceleration of a point on the rim, and (d) the total distance a point our tke rim noves ign-2.00 s (E) The moment of inertia if it's mass is 2Kg? is the answer

5 0
3 years ago
The diagram is an illustration of<br> NO LINKS
V125BC [204]

Answer:

A longitudinal wave

Explanation:

6 0
3 years ago
A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com
bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature =20^{\circ}C

Final Temperature =80^{\circ}C

mass flow rate of cold fluid \dot{m_c}=1.2 kg/s

Initial Geothermal water temperature T_h_i=160^{\circ}C

Let final Temperature be T

mass flow rate of geothermal water \dot{m_h}=2 kg/s

diameter of inner wall d_i=1.5 cm

U_{overall}=640 W/m^2K

specific heat of water c=4.18 kJ/kg-K

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)

2\times (160-T)=1.2\times (80-20)

160-T=36

T=124^{\circ}C

As heat exchanger is counter flow therefore

\Delta T_1=160-80=80^{\circ}C

\Delta T_2=124-20=104^{\circ}C

LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}

LMTD=\frac{80-104}{\ln \frac{80}{104}}

LMTD=91.49^{\circ}C

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

\dot{m_c}c(80-20)=U\cdot A\cdot (LMTD)

A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2

A=\pi DL=5.144

L=\frac{5.144}{\pi \times 0.015}

L=109.16 m

6 0
3 years ago
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