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ivanzaharov [21]

1 year ago

11

Levi and Clara are trying to move a very heavy box. Levi is pushing the box with a force of 30 N, and Clara is pulling the box w

ith a force of 50 N. The box exerts a frictional force of 65 N.
Which best explains the forces?
Answer
There is a net force of 15 N allowing Levi and Clara to move the box.

2 answers:

san4es73 [151]1 year ago

7 0

Answer:

There is a net force of 15 N allowing Levi and Clara to move the box.

Explanation:

Komok [63]1 year ago

5 0

There is a ner force of 15 N allowing Levi and Clara to mobe the box.

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In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec

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Answer with Explanation:

We are given that

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a.We have to find the electric potential of the proton at the position of the electron.

We know that the electric potential

$V=\frac{kq}{r}$

Where $k=9\times 10^9$

$V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$V=27.17 V$

B.Potential energy of electron,U= $\frac{kq_e q_p}{r}$

Where

$q_e=-1.6\times 10^{-19} c=$ Charge on electron

$q_p=q=1.6\times 10^{-19} C$ =Charge on proton

Using the formula

$U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$U=-4.35\times 10^{-18} J$

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When driving in heavy rain, or on a flooded road, your tires can ride on a thin film of water like skis;

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The answer is letter a. It is best to slow down in situations of heavy rain or flooded road as skid could be the result if you lose out of control because the driver isn't slowing down. That is why it is being said that tires can ride on a thin film of water skis as it could skid if it has lost control if the driver hadn't slowed down.

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A rock is thrown straight up with an initial velocity of 19.6 m/s. What time interval elapses between the rock’s being thrown an

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Answer:

It will take 4 sec rock to comes its original point

Explanation:

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1 year ago

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A circular coil 17.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this locati

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Answer:

The torque in the coil is 4.9 × 10⁻⁵ N.m

Explanation:

T = NIABsinθ

Where;

T is the torque on the coil

N is the number of loops = 9

I is the current = 7.8 A

A is the area of the circular coil = ?

B is the Earth's magnetic field = 5.5 × 10⁻⁵ T

θ is the angle of inclination = 90 - 56 = 34°

Area of the circular coil is calculated as follows;

$A = \frac{\pi d^2}{4} \\\\A = \frac{\pi 0.17^2}{4} =0.0227 m^2$

T = 9 × 7.8 × 0.0227 × 5.5×10⁻⁵ × sin34°

T = 4.9 × 10⁻⁵ N.m

Therefore, the torque in the coil is 4.9 × 10⁻⁵ N.m

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1 year ago

for a given initial projectile speed, you observe that the projectile has a certain range R at a launch angle of a = 30. For wha

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Answer:

The other angle is 30 degrees.

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The range of projectile is given by :

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Here,

u is the speed of launch of projectile

Here, $\theta=30^{\circ}$

We need to find the other launch angle when the projectile have the same range, such that,

$\dfrac{u^2\ \sin(60)}{g}=\dfrac{u^2\ \sin2\alpha}{g}$

$\sin(60)=\sin2\alpha$

$\dfrac{\sqrt3}{2}=\sin2\alpha$

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So, the other angle is 30 degrees. Hence, this is the required solution.

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