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Alja [10]
3 years ago
10

A bowler once measured that she can throw the bowling ball with a speed of 15miles/hour.If it takes 3 seconds from the ball to t

ravel from the beginning of the lane down to the pins,how far is this distance in feet?
Physics
1 answer:
Vinvika [58]3 years ago
8 0

The distance traveled by the ball in feet is 66 ft.

The given parameters;

  • speed of the ball, u = 15 miles/hour
  • time of motion of the ball, t = 3 seconds

1 mile = 5280 feet

1 hour = 3600 s

The speed of the ball in feet per second is calculated as follows;

v = \frac{15 \ miles}{hour} \times \frac{5280 \ ft}{1 \ mile} \times \frac{1 \ hour}{3600 \ s} = 22 \ ft/s

The distance traveled by the ball in feet is calculated as follows;

Distance = speed x time

Distance = 22 ft/s x 3 s

Distance = 66 ft

Thus, the distance traveled by the ball in feet is 66 ft.

Learn more here: brainly.com/question/13722298

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3 years ago
How long must a 0.70-mm-diameter aluminum wire be to have a 0.40 a current when connected to the terminals of a 1.5 v flashlight
Natalka [10]
By using Ohm's law, we can find what should be the resistance of the wire, R:
R= \frac{V}{I}= \frac{1.5 V}{0.40 A} =3.75 \Omega

Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
r=35 mm=0.35 \cdot 10^{-3} m
So the area is
A=\pi r^2 = \pi (0.35 \cdot 10^{-3} m)^2=3.85 \cdot 10^{-7} m^2

And by using the resistivity  of the Aluminum, \rho=2.65 \cdot 10^{-8} \Omega m, we can use the relationship between resistance R and resistivity:
R= \frac{\rho L}{A}
to find L, the length of the wire:
L= \frac{RA}{\rho}= \frac{(3.75 \Omega)(3,85 \cdot 10^{-7} m^2)}{2.65 \cdot 10^{-8} \Omega m}=54.48 m
4 0
3 years ago
Calculate the mass 9f the earth, assuring that uts is sphere with radius 6.67×10^6m.​
MA_775_DIABLO [31]

Answer:

6.86 × 10²⁴ kg

Explanation:

The mass of the earth m = density of earth, ρ × volume of earth, V

m = ρV

The density of the earth, ρ = 5515 kg/m³ and since the earth is a sphere, its volume is the volume of a sphere V = 4πr³/3 where r = radius of the earth = 6.67 × 10⁶ m

Since m = ρV

m = ρ4πr³/3

So, substituting the values of the variables into the equation for the mass of the earth, m, we have

m = 5515 kg/m³ × 4π(6.67 × 10⁶ m)³/3

m = 5515 kg/m³ × 4π × 296.741 × 10¹⁸ m³/3

m = 5515 kg/m³ × 1189.9639π × 10¹⁸ m³/3

m = 6546105.64378π × 10¹⁸ kg/3

m = 20565197.400122 × 10¹⁸ kg/3

m = 6855065.8 × 10¹⁸ kg

m = 6.8550658 × 10²⁴ kg

m ≅ 6.86 × 10²⁴ kg

8 0
2 years ago
The 0.5kg soccer ball moves toward the net with a force of 4N. What is its acceleration?​
Eduardwww [97]

Answer:

8 m/s²

Explanation:

Given,

Force ( F ) = 4 N

Mass ( m ) = 0.5 kg

To find : -

Acceleration ( a ) = ?

Formula : -

F = ma

a = F / m

= 4 / 0.5

= 40 / 5

a = 8 m/s²

It's acceleration is 8 m/s².

6 0
2 years ago
An astronaut takes an object to the moon where there is less gravity. Explain how the mass and weight of the object on the moon
alexandr1967 [171]
The mass of the object will remain the same rather it's on the moon or on the Earth and even in other places. But the weight will change on the moon, so its weight will be different from the one it had on Earth
3 0
3 years ago
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