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2 years ago

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

1 answer:

7 0

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 metal sphere of radius 10 cm carries a charge of +2.0 μC uniformly distributed over its surface. What is the magnitude of the

frozen [14]

Answer:

$8.0\cdot 10^5 N/C$

Explanation:

Outside the sphere's surface, the electric field has the same expression of that produced by a single point charge located at the centre of the sphere.

Therefore, the magnitude of the electric field ar r = 5.0 cm from the sphere is:

$E=k\frac{q}{(R+r)^2}$

where

$k=8.99\cdot 10^9 N m^2C^{-2}$ is the Coulomb's constant

$q=2.0 \mu C=2.0 \cdot 10^{-6}C$ is the charge on the sphere

$R=10 cm = 0.10 m$ is the radius of the sphere

$r=5.0 cm = 0.05 m$ is the distance from the surface of the sphere

Substituting, we find

$E=(8.99\cdot 10^9 Nm^2 C^{-2})\frac{2.0\cdot 10^{-6} C}{(0.10 m+0.05 m)^2}=8.0\cdot 10^5 N/C$

3 0

2 years ago

A radioactive substance decays exponentially. A scientist begins with 200 milligrams of a radioactive substance. After 17 hours,

lianna [129]

75.17 mg of the radioactive substance will remain after 24 hours.

Answer:

Explanation:

Any radioactive substance will obey the exponential decay behavior. So according to this behavior, any radioactive substance will be decaying in terms of exponential form of disintegration constant and Time.

Disintegration constant is the rate of decay of radioactive elements. It can be measured using the half life time of the radioactive element .While half life time is the time taken by any radioactive element to decay half of its concentration. Like in this case, at first the scientist took 200 mg then after 17 hours, it got reduced to 100 g. So the half life time of this element is 17 hours.

Then Disintegration constant = 0.6932/Half Life time

Disintegration constant = 0.6932/17=0.041

Then as per the law of disintegration constant:

$N = N_{0}e^{-xt}$

Here N is the amount of radioactive element remaining at time t and $N_{0}$ is the initial amount of sample, x is the disintegration constant.

So here, $N_{0}$ = 200 mg, x = 0.041 and t = 24 hrs.

N = 200 × $e^{-24*0.041}$ =75.17 mg.

So 75.17 mg of the radioactive substance will remain after 24 hours.

3 0

1 year ago

An object begins at position x = 0 and moves one-dimensionally along the x-axis witļi a velocity v

Liula [17]

Answer:

The answer is "between 20 s and 30 s".

Explanation:

Calculating the value of positive displacement:

$\ (x_{+ve}) = \frac{1}{2} \times 15 \times 20 \\\\$

$= \frac{1}{2} \times 300 \\\\= 150 \\\\$

Calculating the value of negative displacement upon the time t:

$(x_{-ve}) = \frac{1}{2} \times 5 \times 20- 20(t-20) \\\\$

$= \frac{1}{2} \times 100- 20t+ 400 \\\\= 50- 20t+ 400 \\\\$

$\to X= X_{+ve} + X_{-ve} \\\\$

$\to 150 - 50 -20t+400 =0\\\\\to 100 -20t+400 =0 \\\\\to 500 -20t =0\\\\\to 20t =500 \\\\\to t=\frac{500}{20}\\\\\to t=\frac{50}{2}\\\\\to t= 25$

That's why its value lie in "between 20 s and 30 s".

6 0

2 years ago

Determine the centripetal force upon a 40-kg child who makes 10 revolution around the cliffhanger in 29.3 seconds.the radius of

zysi [14]

Answer:

The centripetal force acting on the child is 39400.56 N.

Explanation:

Given:

Mass of the child is, $m=40\ kg$

Radius of the barrel is, $R=2.90\ m$

Number of revolutions are, $n =10$

Time taken for 10 revolutions is, $t=29.3\ s$

Therefore, the time period of the child is given as:

$T=\frac{n}{t}=\frac{10}{29.3}=0.341\ s$

Now, angular velocity is related to time period as:

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.341}=18.43\ rad/s$

Now, centripetal force acting on the child is given as:

$F_{c}=m\omega^2 R\\F_{c}=40\times (18.43)^2\times 2.90\\F_{c}=40\times 339.66\times 2.90\\F_{c}=39400.56\ N$

Therefore, the centripetal force acting on the child is 39400.56 N.

8 0

2 years ago

A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of

olya-2409 [2.1K]

Explanation:

When Michelson-Morley apparatus is turned through $90^{o}$ then position of two mirrors will be changed. The resultant path difference will be as follows.