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3 years ago

11

If you ran 15 km/hr for 2.5 hours, how much distance would you cover?

1 answer:

SVETLANKA909090 [29]3 years ago

6 0

Answer: 37.5 km

Explanation:

The question is that

If you ran 15 km/hr for 2.5 hours, how much distance would you cover ?

Where

Speed = 15 km/ hr

Time = 2.5 hours

Using the formula for speed.

Speed = distance/time

Substitute speed and time into the formula

15 = distance/ 2.5

Make distance the subject of formula by cross multiplying.

Distance = 15 × 2.5

Distance = 37.5 km.

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A frictionless spring with a 3-kg mass can be held stretched 0.8 meters beyond its natural length by a force of 40 newtons. If t

Free_Kalibri [48]

Answer:

Explanation:

mass m = 3 kg

spring constant be k

k x .8 = 40 N

k = 40 / .8 = 50 N /m

angular frequency ω = √ ( k / m )

= √ ( 50 / 3 )

= 4.08 rad /s

Let amplitude of oscillation be A .

1/2 k A² = 1/2 m v²

50 A² = 3 x 1²

A = .245 m = 24.5 cm

For displacement , the equation of SHM is

x = A sinωt

= 24.5 sin4.08 t

x = 24.5 sin4.08 t

Here, angle 4.08 t is in radians .

3 0

3 years ago

A 980 kg roller coaster cart is traveling along a track at 17 m/s before it rolls down a 30 m tall hill (Point A). What will be

MrRissso [65]

The kinetic energy halfway the hill is $2.86\cdot 10^5 J$

Explanation:

If there are no friction forces acting on the cart, we can apply the law of conservation of energy: the mechanical energy of the cart (which is the sum of potential energy + kinetic energy) must be conserved. So we can write:

$U_A +K_A = U_B + K_B$

where

$U_A=mgh_A$ is the initial potential energy, at point A, with

m = 980 kg (mass of the cart)

$g=9.8 m/s^2$ (acceleration of gravity)

$h_A = 30 m$ (height at point A)

$K_A=\frac{1}{2}mv_A^2$ is the initial kinetic energy, at point A , with

$v_A=17 m/s$ (velocity at point A)

$U_B=mgh_B$ is the final potential energy, at point B, where

$h_B = 15 m$ (height at point B)

$K_B=\frac{1}{2}mv_B^2$ is the final kinetic energy, at point B, where

$v_B$ is the velocity at point B

Here we are interested in finding $K_B$ , so by re-arranging the equation and substituting we find:

$K_B = U_A+K_B-U_B = mg(h_A-h_B)+\frac{1}{2}mv_A^2=(980)(9.8)(30-15)+\frac{1}{2}(980)(17)^2=2.86\cdot 10^5 J$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

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3 years ago

A cup of coffee is sitting on a table in a train that is moving with a constant velocity. The coefficient of static friction bet

Vikki [24]

Answer:

a = 2.94 m/s²

Explanation:

In order for the cup not to slip, the unbalanced force on cup must be equal to the frictional force:

Unbalanced Force = Frictional Force

ma = μR = μW

ma = μmg

a = μg

where,

a = maximum acceleration for the cup not to slip = ?

μ = coefficient of static friction = 0.3

g = acceleration due to gravity = 9.8 m/s²

Therefore,

a = (0.3)(9.8 m/s²)

<u>a = 2.94 m/s²</u>

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3 years ago

DESCRIBE THE REQUIREMENTS OF AN INTERNET CONNECTION?<br>please tell me the answer

AlekseyPX

Answer: The basic requirements for connecting to the Internet are a computer device, a working Internet line, and the right modem for that Internet line. In addition, software programs such as Internet browsers, email clients, Usenet clients, and other special applications are needed in order to access the Internet.

Explanation: brainleist pls :)

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3 years ago

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Whitepunk [10]

Answer:

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Explanation:

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3 years ago

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