All categories

3 years ago

Jenna can run 9 km in 1 hour. If she runs at that speed for 30 minutes, how far will she go? *

Physics

1 answer:

crimeas [40]3 years ago

5 0

Answer:

4.5 km / 450 m

Explanation:

1 hour = 9 km

or, 60 min = 9km

or, 60/2 min = 9/2 km

■ 30 min = 4.5 km

heart plzz and brainliest answer as well

You might be interested in

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has

irga5000 [103]

Answer:

The total linear acceleration is approximately 0.246 meters per square second.

Explanation:

The total linear acceleration ( $a$ ) consist in two components, <em>radial</em> ( $a_{r}$ ) and <em>tangential</em> ( $a_{t}$ ), in meters per square second:

$a_{r} = \omega^{2}\cdot r$ (1)

$a_{t} = \alpha \cdot r$ (2)

Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$ (3)

Where:

$r$ - Radius of the wheel, in meters.

$\omega$ - Angular speed, in radians per second.

$\alpha$ - Angular acceleration, in radians per square second.

Given that wheel accelerates uniformly, we use the following kinematic equation:

$\omega = \omega_{o}+ \alpha\cdot t$ (4)

Where:

$\omega_{o}$ - Initial angular speed, in radians per second.

$t$ - Time, in seconds.

If we know that $r = 0.1\,m$ , $\alpha = 2\,\frac{rad}{s^{2}}$ , $\omega_{o} = 0\,\frac{rad}{s}$ and $t = 0.60\,s$ , then the total linear acceleration is:

$\omega = \omega_{o}+ \alpha\cdot t$

$\omega = 1.2\,\frac{rad}{s}$

$a_{r} = \omega^{2}\cdot r$

$a_{r} = 0.144\,\frac{m}{s^{2}}$

$a_{t} = \alpha \cdot r$

$a_{t} = 0.2\,\frac{m}{s^{2}}$

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$

$a \approx 0.246\,\frac{m}{s^{2}}$

The total linear acceleration is approximately 0.246 meters per square second.

3 0

3 years ago

A point charge is placed 100 m from a 6 uC charge generating an electric field. What is the

kvv77 [185]

The strength of the electric field is 5 N/C

Explanation:

The magnitude of the electric field produced by a single-point charge is given by:

$E=\frac{kQ}{r^2}$

where

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

Q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

$Q=6 \mu C = 6\cdot 10^{-6}C$ is the charge producing the field

r = 100 m is the distance from the charge at which we want to calculate the field

Substituting into the equation, we find the s trength of the electric field:

$E=\frac{(8.99\cdot 10^9)(6\cdot 10^{-6})}{(100)^2}=5.4 N/C \sim 5 N/C$

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

3 0

3 years ago

What are the characteristics of an ideal transformer

Vsevolod [243]

It is an imaginary transformer which has no core loss, no ohmic resistance and no leakage flux. The ideal transformer has the following important characteristic. The resistance of their primary and secondary winding becomes zero. The core of the ideal transformer has infinite permeability.

4 0

3 years ago

PLEASE ANSWER ASAP

neonofarm [45]

Answer:

2.55sec

Explanation:

time = distance/speed = (87 m)/(34 m/s) = 2.55sec

6 0

2 years ago

Which states of matter have particles that move independently of one another with very little attraction

Crazy boy [7]

The state of matter that the particles move independently of one another with very little attraction is, I believe, gas

6 0

3 years ago