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baherus [9]
3 years ago
14

If you lift two loads up one story, how much work do you do compared to lifting just one load up one story?

Physics
1 answer:
Daniel [21]3 years ago
6 0

Answer:a

Explanation:

We have to lift the load two loads up one story, so energy required is

let m be the mass of each load and h is the height of each story

Energy E_1=2\times m\times g\times \h

E_1=2mgh

Here energy gained is the potential energy which depends upon the datum(floor).

For lifting one load up one story

Energy required

E_2=mgh

thus E_2 is half of E_1

So option a is correct                                

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A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x (F
Marrrta [24]

Answer:

Part a)

x = 0.4 m

Part b)

v_i = 11.7 m/s

Part c)

Speed is more than the required speed so it will reach the top

Explanation:

Part a)

As we know that there is no frictional force while block is moving on horizontal plane

so we can use energy conservation on the block

\frac{1}{2}mv^2 = \frac{1}{2}kx^2

\frac{1}{2}0.500(12^2) = \frac{1}{2}(450)x^2

x = 0.4 m

Part b)

If the track has average frictional force of 7 N then work done by friction while block slides up is given as

W_f = -7( \pi R)

W_f = -7(\pi \times 1.00)

W_f = -22 J

work done against gravity is given as

W_g = - mg(2R)

W_g = -(0.500)(9.8)(2\times 1)

W_g = -9.8 J

Now by work energy equation we have

\frac{1}{2}mv_i^2 + W_f + W_g = \frac{1}{2}mv_f^2

\frac{1}{2}0.5(12^2) - 9.8 - 22 = \frac{1}{2}(0.5)v_f^2

v_f = 4.1 m/s

Part c)

now minimum speed required at the top is such that the normal force must be zero

mg = \frac{mv^2}{R}

v = \sqrt{Rg}

v = 3.13 m/s

so here we got speed more than the required speed so it will reach the top

5 0
3 years ago
A proton and an electron are moving in the +x direction in a magnetic field in the +z
Cloud [144]

Answer:

Explanation:

A proton and electron are moving in the positive x direction, this shows that their velocity will be in the positive x direction

V = v•i

Magnetic field Is the positive z direction

B = B•k

A. For proton.

Proton has a positive charge of q

Direction of force on proton

Force is given as

F = q(v×B)

F = q( v•i × B•k)

F = qvB (i×k)

From vectors i×k = -j

F = -qvB •j

Then, for the positive charge, the force will act in the negative direction of the y-axis

B. For electron

Electron has a negative of -q

Direction of force on proton

Force is given as

F = q(v×B)

F = -q( v•i × B•k)

F = -qvB (i×k)

From vectors i×k = -j

F = --qvB •j

F = qvB •j

Then, for the negative charge, the force will act in the positive direction of the y-axis

5 0
3 years ago
If a person walks 3 m north and 5 meters east, how would you find the displacement for that person? what would the displacement
UkoKoshka [18]

Answer:

AC)=(AB)2+(BC)2−−−−−−−−−−−−√=42+32−−−−−−√

⇒displacement=16+9−−−−−√=25−−√=5m

8 0
3 years ago
Which best describe insane
mash [69]

in a state of mind that prevents normal perception, behavior, or social interaction; seriously mentally il

4 0
3 years ago
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There is strong evidence that Europa, a satellite of Jupiter, has a liquid ocean beneath its icy surface. Many scientists think
dangina [55]

Answer:

4.44 rpm

Explanation:

\omega = Angular speed

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Europa = \frac{3138000}{2}\ m

R = Radius of arm = 6 m

The acceleration due to gravity is given by

g=\frac{GM}{r^2}\\\Rightarrow g=\frac{6.67\times 10^{-11}\times 4.8\times 10^{22}}{\left(\frac{3138000}{2}\right)^2}\\\Rightarrow g=1.3\ m/s^2

Here the centripetal acceleration of the arm and acceleration due to gravity are equal

a_c=\omega^2R

a_c=g\\\Rightarrow \omega^2R=1.3\\\Rightarrow \omega^2\times 6=1.3\\\Rightarrow \omega=\sqrt{\frac{1.3}{6}}\\\Rightarrow \omega=0.46547\ rad/s

Converting to rpm

1\ rad/s=\frac{60}{2\pi}\ rpm

0.46547\ rad/s=0.46547\times \frac{60}{2\pi}\ rpm=4.44\ rpm

The angular speed of the arm is 4.44 rpm

8 0
3 years ago
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