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2 years ago

4. How much time does it take for a student running at a speed of 5 m/s to cover a distance of 2,000 m?

Physics

1 answer:

Monica [59]2 years ago

6 0

Answer:

6 Minutes 40 Seconds or 400 Seconds

Explanation:

Time to cover a distance of 5m = 1 Second

Time to cover a distance of 2000m = 2000÷5

= 400 Seconds

After converting 400 Seconds into minutes it will become 6 minutes 40 seconds.

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A 4.40-m-long, 500 kg steel uniform beam extends horizontally from the point where it has been bolted to the framework of a new

icang [17]

Answer:

Torque=13798.4 N.m

Explanation:

Given data

Mass of beam m₁=500 kg

Mass of the person m₂=70 kg

length of steel r₁=4.40m

center of gravity of the beam is at r₂=r₁/2 =4.40/2 = 2.20m

To find

Torque

Solution

Torque due to beam own weight

$T_{1}=m_{1}gr_{1}\\ T_{1}=500*2.2*9.8\\T_{1}=10780N.m$

Torque due to person

$T_{2}=m_{2}r_{2}g\\T_{2}=70*(4.40)*(9.8)\\T_{2}=3018.4 N.m$

Now for total torque

$T_{total}=T_{1}+T_{2}\\T_{total}=10780+3018.4\\T_{total}=13798.4N.m$

4 0

3 years ago

Problem 21-40a:

Greeley [361]

Answer:

Part a)

$\frac{F_e}{F_g} = 2.74 \times 10^{12}$

Part b)

$q = 3.37 \times 10^{-4} C$

Explanation:

As we know that electric force on electric charge is given as

$F = qE$

here we have

$q = 1.6 \times 10^{-19}C$

E = 153 N/C

now force is given as

$F = (1.6 \times 10^{-19})(153) = 2.45 \times 10^{-17} N$

Gravitational force on electric charge near surface of earth is given as

$F_g = mg$

$F_g = (9.1 \times 10^{-31})(9.81) = 8.93 \times 10^[-30} N$

now the ratio of two forces is given as

$\frac{F_e}{F_g} = \frac{2.45 \times 10^{-17}}{8.93 \times 10^{-30}}$

$\frac{F_e}{F_g} = 2.74 \times 10^{12}$

Part b)

Now the ball is balanced by the electric force and the force of gravity on it

so here we have

$F_g = qE$

$mg = qE$

$(5.25 \times 10^{-3})(9.81) = q(153)$

here we have

$q = 3.37 \times 10^{-4} C$

8 0

3 years ago

The half-life of the radioactive element beryllium-13 is 5 × 10-10 seconds, and half-life of the radioactive element beryllium-1

telo118 [61]

<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>

Explanation:

The half-life $h$ of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.

In this case, we are given the half life of two elements:

beryllium-13: $h_{B-13}=5(10)^{-10}s=0.0000000005s$

beryllium-15: $h_{B-15}=2(10)^{-7}s=0.0000002s$

As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?

We can find it out by the following expression:

$h_{B-15}=X.h_{B-13}$

Where $X$ is the amount we want to find:

$X=\frac{h_{B-15}}{h_{B-13}}$

$X=\frac{2(10)^{-7}s}{5(10)^{-10}s}$

Finally:

$X=400$

Therefore:

The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.

8 0

2 years ago

Which feature of the sun appears in cycles of about 11 years?

goldfiish [28.3K]

The Sun's magnetic field goes through a cycle, called the solar cycle. Every 11 years or so, the Sun's magnetic field completely flips. This means that the Sun's north and south poles switch places. Then it takes about another 11 years for the Sun's north and south poles to flip back again.

8 0

3 years ago

HELPPP WILL MARK BRAINLIST What type of image is formed by a lens if m= 2.7?

Gelneren [198K]

Answer:

the answer is C

Explanation:

So, the formed image will be real and inverted.

6 0

2 years ago