Ask question

Ask question

All categories

3 years ago

13

A 1 250 kg car moving at a velocity of 30 km/hr along EDSA is accelerated by a force of 1 700 N. What will be its velocity after

10 seconds? (Neglect friction)

1 answer:

Talja [164]3 years ago

6 0

<h3><u>The velocity of the car after 10 s is 78.95 km/hr</u></h3>

Explanation:

<h2>Given:</h2>

m = 1,250 kg

$v_i$ = 30 km/hr

F = 1,700 N

t = 10 s

<h2>Required:</h2>

Final velocity

<h2>Equation:</h2><h3>Force</h3>

F = ma

where: F - force

m - mass

a - acceleration

<h3>Acceleration</h3>

a = $\frac{v_f \:-\:v_i}{t}$

where: a - acceleration

$v_i$ - initial velocity

$v_f$ - final velocity

t - time elapsed

<h2>Solution:</h2><h3>Solve for acceleration using the formula for force</h3>

F = ma

Substitute the value of F and m

(1700 N) = (1250 kg)(a)

a = $\frac{1700\:N}{1250\:N}$

a = 1.36 m/s²

<h3>Solve for final velocity using the formula for acceleration</h3>

Convert 30 km/hr to m/s

= $\frac{30\:km}{hr}\:×\:\frac{1000\:m}{1\:m}\:×\:\frac{1\:hr}{3600\:s}$

= $8.33 m/s$

Substitute the value of a, $v_i$ and t

a = $\frac{v_f \:-\:v_i}{t}$

$1.36\: m/s² \:= \:\frac{v_f \:-\:8.33\:m/s}{10\:s}$

$(10 \:s)1.36\: m/s² \:= \:v_f \:-\:8.33\:m/s$

$v_f\: =\: (10 \:s)1.36 \:m/s²\: + \:8.33\:m/s$

$v_f \: =\: 13.6 \:m/s \:+\: 8.33\:m/s$

$v_f\: =\: 21.93\: m/s$

Convert to km/hr

= $\frac{21.93\:m}{s}\:×\:\frac{1\:km}{1000\:m}\:×\:\frac{3600\:s}{1/:hr}$

= $78.95\: km/hr$

<h2>Final answer</h2><h3><u>The velocity of the car after 10 s is 78.95 km/hr</u></h3>

You might be interested in

The 1000-lb elevator is hoisted by the pulley system and motor M. The motor exerts a constant force of 500 lb on the cable. The

klemol [59]

The power that must be supplied to the motor is 136 hp

<u>Explanation:</u>

Given-

weight of the elevator, m = 1000 lb

Force on the table, F = 500 lb

Distance, s = 27 ft

Efficiency, ε = 0.65

Power = ?

According to the equation of motion:

F = ma

$3(500) - 1000 = \frac{1000}{32.2} * a$

a = 16.1 ft/s²

We know,

$v^2 - u^2 = 2a (S - So)\\\\v^2 - (0)^2 = 2 * 16.1 (27-0)\\\\v = 29.48m/s$

To calculate the output power:

Pout = F. v

Pout = 3 (500) * 29.48

Pout = 44220 lb.ft/s

As efficiency is given and output power is known, we can calculate the input power.

ε = Pout / Pin

0.65 = 44220 / Pin

Pin = 68030.8 lb.ft/s

Pin = 68030.8 / 500 hp

= 136 hp

Therefore, the power that must be supplied to the motor is 136 hp

5 0

4 years ago

Who plays a role in the financial activities of a company?

KatRina [158]

Hey,

Who plays a role in the financial activities of a company?

<em>O D. Everyone at the company, including managers and employees</em>

3 0

4 years ago

Salvage ethnography is the effort to ensure that ethnography remains an important part of anthropology. recording of linguistic

fenix001 [56]

Answer:

D

Explanation:

D

6 0

4 years ago

A ceramic matrix composite contains internal flaws as large as 0.001 cm in length. The plane strain fracture toughness of the co

murzikaleks [220]

Since the applied stress required for failure due to crack propagation is still higher than 550 MPa, the ceramic is expected to fail due to overload and not because of the flaws

Explanation:

<u>Plane -Strain Fracture toughness is calculated as</u>

$k_{IC}$ = $f$ б $\sqrt{\pi a}$

F=geometry factor of the flaw

б=Stress applied

$k_{IC}$ =Fracture toughness

a=Flaw size

<u>Given that </u>

Internal Flaw,a=0.001cm

Fracture Toughness $k_{IC}$ =45MPa $\sqrt{m}$

Tensile Strength б=550 MPa

Geometry Factor, $f$ =1

<u>Calculation</u>

An internal Flaw i s 0.001 cm

2a=0.001cm

a=0

6 0

4 years ago

A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the

raketka [301]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

(c) change in torque angle = ?

(c) rotor speed = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

$E = G \times H$

Where G represents complex rated power and H is the inertia constant of turbo-generator.

$E = 100 \times 8 \\\\E = 800 \: MJ$

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

$P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}$

Where M is given by

$M = \frac{E}{180 \times f}$

$M = \frac{800}{180 \times 50}$

$M = 0.0889 \: MJ \cdot s/ elec \: \: deg$

So, the rotor acceleration is

$P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}$

$30 = 0.0889 \frac{d^2 \delta}{dt^2}$

$\frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}$

$\frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2$

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

$\Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2$

Where t is given by

$1 \: cycle = 1/f = 1/50 \\\\10 \: cycles = 10/50 = 0.2 \\\\t = 0.2 \: sec$

So,

$\Delta \delta = \frac{1}{2} \cdot 337.46 \cdot (0.2)^2$

$$ \Delta \delta = 6.75 \: elec \: deg$

The change in torque in rpm/s is given by

$\Delta \delta = \frac{337.46 \cdot 60}{2 \cdot 360\circ }$

$\Delta \delta =28.12 \: \: rpm/s$

The rotor speed in revolutions per minute at the end of this period (10 cycles) is given by

$Rotor \: speed = \frac{120 \cdot f}{P} + (\Delta \delta)\cdot t$

Where P is the number of poles of the turbo-generator.

$Rotor \: speed = \frac{120 \cdot 50}{4} + (28.12)\cdot 0.2$

$Rotor \: speed = 1500 + 5.62$

$$ Rotor \: speed = 1505.62 \:\: rpm$

4 0

4 years ago