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3 years ago

A runner runs 4875 ft in 6.85 minutes. what is the runners average speed in miles per hour?

Physics

2 answers:

yanalaym [24]3 years ago

8 0

The average speed can be easily calculated by taking the ratio of distance and time. That is:

average speed = distance / time

so calculating:

average speed = 4875 ft / 6.85 minutes

average speed = 711.68 ft / min

zaharov [31]3 years ago

7 0

Answer:

Speed of the runner is 8.08 miles per hour approximately.

Explanation:

Given:

A runner runs =4875 ft

Time Taken = 6.85 minutes.

To Find:

Average speed of the runner in miles per hour=?

Solution:

Distance = Speed x time

Conveting feet into miles:

=>Distance= $4875\times\frac{1}{5280}$ miles

=>Distance = 0.9232955 miles

Converting minutes into hours:

time = 6.85 minutes

=> $6.85\times \frac{1}{60}$ hours

=> $\frac{6.85}{60}$ = 0.1141667 hours.

Now, we know that, distance = speed x time

Then, 0.9232955 miles = speed x 0.1141667 hours

=>speed = $\frac{0.9232955}{0.1141667}$

=>speed = 8.08725749

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Alpha particles, because they are the heaviest ones (helium nuclei) and will travel around the body.

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3 years ago

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

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3 years ago

Water being turned into ice cubes in a freezer is an example of _____.

Gekata [30.6K]

Answer:

a physical change

Explanation:

after the water turns to ice, it will melt and became water again making which means it's reversible this being. a physical change

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2 years ago

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A wave is traveling at 241 m/s. It has a wavelength of 30.0 m. What is its frequency?

Feliz [49]

Answer:

A) 8.03Hz

Explanation:

f = V/λ

Where wavelength( λ )= 30m

Speed (V) =241m/S

f= 241/30=8.03Hz

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3 years ago

A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the

lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

$l^2 = x^2 + y^2-----(1)$