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2 years ago

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A car is moving along a straight line op is shown below it moves from O to P in 8sec and return to p to q in 6sec. What is the a

verage velocity and the speed of the car

1 answer:

Pavel [41]2 years ago

4 0

Answer:

The answer would be A

Explanation:

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High-speed stroboscopic photographs show that the head of a 210-g golf club is traveling at 56 m/s just before it strikes a 46-g

tamaranim1 [39]

Explanation:

It is given that,

Mass of golf club, m₁ = 210 g = 0.21 kg

Initial velocity of golf club, u₁ = 56 m/s

Mass of another golf ball which is at rest, m₂ = 46 g = 0.046 kg

After the collision, the club head travels (in the same direction) at 42 m/s. We need to find the speed of the golf ball just after impact. Let it is v.

Initial momentum of golf ball, $p_i=m_1u_1=0.21\ kg\times 56\ m/s=11.76\ kg-m/s$

After the collision, final momentum $p_f=0.21\ kg\times 42\ m/s+0.046v$

Using the conservation of momentum as :

$p_i=p_f$

$11.76\ kg-m/s=0.21\ kg\times 42\ m/s+0.046v$

v = 63.91 m/s

So, the speed of the golf ball just after impact is 63.91 m/s. Hence, this is the required solution.

3 0

3 years ago

d is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy is J?

Anna71 [15]

<span>D is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy is J?
</span>Answer: The sled's potential energy is 882 Joules

5 0

3 years ago

Read 2 more answers

A spring stretches 0.018 m when a 2.8-kg object is suspendedfrom its end. How much mass should be attached to this spring sothat

agasfer [191]

Answer:

m = 4.29 kg

Explanation:

Given that,

Mass of the object, m = 2.8 kg

Stretching in the spring, x = 0.018 m

Frequency of vibration, f = 3 Hz

Let m is the mass of the object that is attached to the spring. When it is attached the gravitational force is balanced by the force on spring. It is given by :

$mg=kx$

$k=\dfrac{mg}{x}$

$k=\dfrac{2.8\times 9.8}{0.018}$

k = 1524.44 N/m

Since, $\omega=\sqrt{\dfrac{k}{m}}$

$m=\dfrac{k}{4\pi^2 f^2}$

$m=\dfrac{1524.44}{4\pi^2 \times 3^2}$

m = 4.29 kg

So, the mass that is attached to this spring is 4.29 kg. Hence, this is the required solution.

4 0

3 years ago

The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the

Eddi Din [679]

Answer:

x = D (M/M-m) 2.41

Explanation:

a) Let's apply Newton's second law to find the summation of force, where each force is given by the law of universal gravitation

F = g m₁m₂ / r²

Σ F = 0

F1- F2 = 0

F1 = F2

We set the reference system in the body of greatest mass (M) the planet

F1 = g m₁ M / x²

F2 = G m1 m / (D-x)²

G m₁ M / x² = G m₁ m / (D-x)²

M (D-x)² = m x²

MD² -2MD x + M x² = m x²

x² (M-m) -2MD x + MD² = 0

We solve the second degree equation

x = [2MD ±√ (4M²D² - 4 (M-m) MD²)] / 2 (M-m)

x = {2MD ± 2D √ (M² + (M-m) M)} / 2 (M-m)

x = D {M ± Ra (2M²-mM)} / (M-m)

x = D (M ± M √ (2-m/M)) / (M-m)

x = D (M / (M-m)) (1 ±√ (2-m/M)

Let's analyze this result, the value of M-m >> 1, so if we take the negative root, the value of x would be negative, it is out of the point between the two bodies, so the correct result must be taken with the positive root

x = D (M / (M-m)) (1 + √2)

x = D (M/M-m) 2.41

b) X = 2/3 D

x = D (M/M-m) 2.41

2/3 D = D (M/(M-m)) 2.41

2/3 (M-m) = M 2.41

2/3 M - 2/3 m = 2.41 M

1.743 M = 0.667 m

M/m = 0.667/1.743

M/m = 0.38

3 0

3 years ago

Steam enters an adiabatic turbine steadily at 7 MPa, 5008C, and 45 m/s, and leaves at 100 kPa and 75 m/s. If the power output of

marusya05 [52]

Answer:

a) $\dot m = 6.878\,\frac{kg}{s}$ , b) $T = 104.3^{\textdegree}C$ , c) $\dot S_{gen} = 11.8\,\frac{kW}{K}$

Explanation:

a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.

$-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0$

The mass flow rate is:

$\dot m = \frac{\dot W_{out}}{h_{in}-h_{out}}$

According to property water tables, specific enthalpies and entropies are:

State 1 - Superheated steam

$P = 7000\,kPa$

$T = 500^{\textdegree}C$

$h = 3411.4\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

State 2s - Liquid-Vapor Mixture

$P = 100\,kPa$

$h = 2467.32\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

$x = 0.908$

The isentropic efficiency is given by the following expression:

$\eta_{s} = \frac{h_{1}-h_{2}}{h_{1}-h_{2s}}$

The real specific enthalpy at outlet is:

$h_{2} = h_{1} - \eta_{s}\cdot (h_{1}-h_{2s})$

$h_{2} = 3411.4\,\frac{kJ}{kg} - 0.77\cdot (3411.4\,\frac{kJ}{kg} - 2467.32\,\frac{kJ}{kg} )$

$h_{2} = 2684.46\,\frac{kJ}{kg}$

State 2 - Superheated Vapor

$P = 100\,kPa$

$T = 104.3^{\textdegree}C$

$h = 2684.46\,\frac{kJ}{kg}$

$s = 7.3829\,\frac{kJ}{kg\cdot K}$

The mass flow rate is:

$\dot m = \frac{5000\,kW}{3411.4\,\frac{kJ}{kg} -2684.46\,\frac{kJ}{kg}}$

$\dot m = 6.878\,\frac{kg}{s}$

b) The temperature at the turbine exit is:

$T = 104.3^{\textdegree}C$

c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:

$\dot m \cdot (s_{in}-s_{out}) + \dot S_{gen} = 0$

$\dot S_{gen}=\dot m \cdot (s_{out}-s_{in})$

$\dot S_{gen} = (6.878\,\frac{kg}{s})\cdot (7.3829\,\frac{kJ}{kg\cdot K} - 6.8000\,\frac{kJ}{kg\cdot K} )$

$\dot S_{gen} = 11.8\,\frac{kW}{K}$

4 0

3 years ago