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r-ruslan [8.4K]
3 years ago
6

It is easier to overcome load when the load is shifted towards the wheel in a wheelbarrow why ? give reason​

Physics
2 answers:
krok68 [10]3 years ago
4 0

Answer:

The wheelbarrow's wheel and axle help the wheelbarrow to move without friction thus making it easier to push or pull. That's why it will be easier to lift a load in wheel barrow of the load is transferred towards the wheel.

mote1985 [20]3 years ago
3 0

Answer:

Replace the wheelbarrow with a plank with one end on the ground.

Place the load L at 1/2 the distance to the other end and lift the other end at the length of the plank. So you only lift a weight of L/2. The wheel also supports a weight of L/2.

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A series RCL circuit consists of a 50 Ω resistor, 500 Ω capacitor, and a 500 Ω inductor. What is the angle of impedance?
lbvjy [14]

Answer:

Zero

Explanation:

here, the inductive reactance and the capacitive reactance is same, so this is the condition for resonance.

In the condition for resonance,

the circuit and the voltage in the circuit is in the same phase and the impedance of the circuit is minimum which is equal to the resistance of the circuit.

The phase angle is given by

tan\phi =\frac{X_{L}-X_{C}}{R}

Ф = 0

5 0
3 years ago
A single-turn circular loop of wire of radius 65 mm lies in a plane perpendicular to a spatially uniform magnetic field. During
BabaBlast [244]

Answer:

Explanation:

change in flux = no of turns x area of loop x change in magnetic field

= 1 x π 65² x 10⁻⁶ x ( 650 - 350 ) x 10⁻³

= 3.9 x 10⁻³ weber .

rate of change of flux  = change of flux / time

= 3.9 x 10⁻³ / .10

= 39 x 10⁻³ V

= 39 mV .

Since the magnetic flux is directed outside page and it is increasing , induced current will be clockwise so that magnetic field is produced in opposite direction to reduce it , as per Lenz's law.

5 0
3 years ago
X(????) = 5.0???? 2 − 4.0???? 3 m.
Burka [1]

Answer:

JA

Explanation:

s of time, (b) the velocity and acceleration at t = 2.0 s, (c) the time at which the position is a maximum, (d) the time at which the velocity is zero, and (e) the maximum position. Assume all variable and constants are in SI units.

4 0
3 years ago
If you weigh 690 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun an
lana [24]

Answer:

Explanation:

Given that,

Mass of star M(star) = 1.99×10^30kg

Gravitational constant G

G = 6.67×10^−11 N⋅m²/kg²

Diameter d = 25km

d = 25,000m

R = d/2 = 25,000/2

R = 12,500m

Weight w = 690N

Then, the person mass which is constant can be determined using

W =mg

m = W/g

m = 690/9.81

m = 70.34kg

The acceleration due to gravity on the surface of the neutron star is can be determined using

g(star) = GM(star)/R²

g(star) = 6.67×10^-11 × 1.99×10^30 / 12500²

g (star) = 8.49 × 10¹¹ m/s²

Then, the person weight on neutron star is

W = mg

Mass is constant, m = 70.34kg

W = 70.34 × 8.49 × 10¹¹

W = 5.98 × 10¹³ N

The weight of the person on neutron star is 5.98 × 10¹³ N

5 0
3 years ago
A bucket of water of mass 14.0 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.260
ruslelena [56]

Answer:

The tension in the rope is 41.38 N.

Explanation:

Given that,

Mass of bucket of water = 14.0 kg

Diameter of cylinder = 0.260 m

Mass of cylinder = 12.1 kg

Distance = 10.7 m

Suppose we need to find that,

What is the tension in the rope while the bucket is falling

We need to calculate the acceleration

Using relation of torque

\tau=F\times r

I\times\alpha=F\times r

Where, I = moment of inertia

\alpha = angular acceleration

\dfrac{Mr^2}{2}\times\dfrac{a}{r}=F\times r

F=\dfrac{M}{2}a...(I)

Here, F = tension

The force is

F=m(g-a)...(II)

Where, F = tension

a = acceleration

From equation (I) and (II)

\dfrac{M}{2}a=m(g-a)

a=\dfrac{g}{1+\dfrac{M}{2m}}

Put the value into the formula

a=\dfrac{9.8}{1+\dfrac{12.1}{2\times14.0}}

a=6.84\ m/s^2

We need to calculate the tension in the rope

Using equation (I)

F=\dfrac{M}{2}a

Put the value into the formula

F=\dfrac{12.1}{2}\times6.84

F=41.38\ N

Hence, The tension in the rope is 41.38 N.

6 0
3 years ago
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