Can anyone help me plz I m confused with this last question?

un motor electric efectueaza un lucru mecanic de 864 j in 0.5 min.tensiunea la bornele sale este de 12v si este parcurs de un cu

GrogVix [38]

Answer:

96%

Explanation:

To find the values of the motor efficiency you use the following formula:

$E=\frac{P_o}{P_i}100$

P_o: output power = 864J/0.5min=864J/30s=28.8W

P_i: input power = I*V = (3A)(12V) = 36W

By replacing this values you obtain:

$E=\frac{28.8W}{30W}*100=96\%$

hence, the motor efficiency is about 96%

traslation:

Pentru a găsi valorile eficienței motorului, utilizați următoarea formulă:

P_o: putere de ieșire = 864J / 0.5min = 864J / 30s = 28.8W

P_i: putere de intrare = I * V = (3A) (12V) = 36W

Înlocuind aceste valori obțineți:

prin urmare, eficiența motorului este de aproximativ 96%

3 0

3 years ago

A body of mass 2.7 kg makes an elastic collision with another body at rest and continues to move in the original direction but w

kramer

Answer:

a)

1.35 kg

b)

2.67 ms⁻¹

Explanation:

a)

$m_{1}$ = mass of first body = 2.7 kg

$m_{2}$ = mass of second body = ?

$v_{1i}$ = initial velocity of the first body before collision = $v$

$v_{2i}$ = initial velocity of the second body before collision = 0 m/s

$v_{1f}$ = final velocity of the first body after collision =

using conservation of momentum equation

$m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}\\(2.7) v + m_{2} (0) = (2.7) (\frac{v}{3} ) + m_{2} v_{2f}\\(2.7) (\frac{2v}{3} ) = m_{2} v_{2f}\\v_{2f} = \frac{1.8v}{m_{2}}$

Using conservation of kinetic energy

$m_{1} v_{1i}^{2}+ m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2} \\(2.7) v^{2} + m_{2} (0)^{2} = (2.7) (\frac{v}{3} )^{2} + m_{2} (\frac{1.8v}{m_{2}})^{2} \\(2.7) = (0.3) + \frac{3.24}{m_{2}}\\m_{2} = 1.35$

b)

$m_{1}$ = mass of first body = 2.7 kg

$m_{2}$ = mass of second body = 1.35 kg

$v_{1i}$ = initial velocity of the first body before collision = 4 ms⁻¹

$v_{2i}$ = initial velocity of the second body before collision = 0 m/s

Speed of the center of mass of two-body system is given as

$v_{cm} = \frac{(m_{1} v_{1i} + m_{2} v_{2i})}{(m_{1} + m_{2})}\\v_{cm} = \frac{((2.7) (4) + (1.35) (0))}{(2.7 + 1.35)}\\\\v_{cm} = 2.67$ ms⁻¹

8 0

3 years ago

Assuming typical speeds of 8.5 km/s and 5.5 km/s for P and S waves, respectively, how far away did the earthquake occur if a par

taurus [48]

Answer:

The earthquake occurred at a distance of 1122 km

Explanation:

Given;

speed of the P wave, v₁ = 8.5 km/s

speed of the S wave, v₂ = 5.5 km/s

The distance traveled by both waves is the same and it is given as;

Δx = v₁t₁ = v₂t₂

let the time taken by the wave with greater speed = t₁

then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.

v₁t₁ = v₂t₂

v₁t₁ = v₂(t₁ + 1.2 min)

v₁t₁ = v₂(t₁ + 72 s)

v₁t₁ = v₂t₁ + 72v₂

v₁t₁ - v₂t₁ = 72v₂

t₁(v₁ - v₂) = 72v₂

$t_1 = \frac{72v_2}{v_1-v_2}\\\\t_1 = \frac{72*5.5}{8.5-5.5}\\\\t_1 = 132 \ s$

The distance traveled is given by;

Δx = v₁t₁

Δx = (8.5)(132)

Δx = 1122 km

Therefore, the earthquake occurred at a distance of 1122 km

4 0

3 years ago

Rohith has been jogging to the bus stop for 2.0 minutes at 3.5 m/s when he looks at

nikdorinn [45]

Multiply these numbers and there’s your answer

6 0

3 years ago

A steel block has a volume of 0.08 m³ and a density of 7,840 kg/m³. What is the force of gravity acting on the block (the weight

Alex17521 [72]

Force of gravity =mass*graviational acceleration
gravitational acceleration=g=9.81
mass=Density*Volume=.08*7840
force of gravity= .08*7840*9.81
gg

5 0

3 years ago