All categories

2 years ago

Nancy ran a distance of 5km in 30 minutes. What is her speed in meters(m) per hour)h)

Physics

1 answer:

Virty [35]2 years ago

7 0

The speed of Nancy who ran a distance of 5 km in 30 minutes is 10000 meters(m) per hour(h).

<h3>What is the rate of speed?</h3>

The rate of speed is the rate at which the total distance is travelled in the time taken. Rate of speed can be given as,

$s=\dfrac{d}{t}$

Here, (d) is the distance travelled by the object and (t) is the time taken but the object to cover that distance.

Nancy ran a distance of 5 km in 30 minutes. There are 60 minutes in one hour. Thus, the time in hour for which Nancy ran is,

$t=\dfrac{30}{60}\\t=0.5\rm\; hr$

The meters in 5 kilometers is,

$d=5\times1000\\d=5000\rm\; m$

She ran 5000 kilometers in 0.5 hours. Thus, the speed of her is,

$s=\dfrac{5000}{0.5}\\s=10000\rm\; km/hr$

Thus, the speed of Nancy who ran a distance of 5 km in 30 minutes is 10000 meters(m) per hour(h).

Learn more about the rate of speed here:

brainly.com/question/359790

#SPJ4

You might be interested in

During oxidation-reduction reactions, a material loses electrons to the oxidizer, which has a positive charge. Which power sourc

Nikolay [14]

Answer:

D. Battery

During the oxidation-reduction(redox) reaction, there is always flow of electrons from one point to another. The electrons are then converted to power through the battery which converts chemical energy to electrical energy. If there is zero flow of electrons then there will also be zero power.

3 0

4 years ago

Read 2 more answers

An open organ pipe 30 cm long and a closed organ pipe 23 cm long, both of same diameter , are each sounding its first overtone ,

cupoosta [38]

First overtone of open organ pipe is given as

$f_{1o} = \frac{v}{L_1 + 2e}$

first overtone of closed organ pipe is given as

$f_{1c} = \frac{3v}{4(L_2 + e)}$

now they are in unison so we will have

$\frac{v}{L_1 + 2e} = \frac{3v}{4(L_2 + e)}$

$\frac{1}{30 + 2e} = \frac{3}{4(23 + e)}$

$90 + 6e = 92 + 4e$

$e = 1 cm$

so end correction of both pipes is e = 1 cm

8 0

3 years ago

You have designed a machine that requires 1000 J of work from a motor for every 800 J of useful work done by the machine. What i

valentinak56 [21]

Answer:

80%

Explanation:

$efficiency = \frac{useful \: work \: done}{total \: energy \: input}$

800 / 1000 = 0.8

Efficiency = 0.8 *100 = 80%

4 0

2 years ago

A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it

makkiz [27]

Answer:D

Explanation:

Given

mass of object $m=5 kg$

Distance traveled $h=10 m$

velocity acquired $v=12 m/s$

conserving Energy at the moment when object start falling and when it gains 12 m/s velocity

Initial Energy $=mgh=5\times 9.8\times 10=490 J$

Final Energy $=\frac{1}{2}mv^2+W_{f}$

$=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}$

where $W_{f}$ is friction work if any

$490=360+W_{f}$

$W_{f}=130 J$

Since Friction is Present therefore it is a case of Open system and net external Force is zero

An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .

4 0

3 years ago

A particle of charge 2.3 ✕ 10−8 C experiences an upward force of magnitude 4.6 ✕ 10−6 N when it is placed in a particular point

Marysya12 [62]

<h2>Answer:</h2>

(a) +2 x 10² N/C (upwards)

(b) -2.2μN or -2.2 x 10⁻⁶N (downwards)

<h2>Explanation:</h2>

The force (F) acting on a particle of charge (Q) at a particular point is related to its electric field (E) by the following;

F = Q x E ----------------------(i)

This means that the force acting on the charged particle is the product of its charge and the electric field at that point.

<em>(a) Given</em>;

Q = charge of the particle = 2.3 x 10⁻⁸ C

F = force acting on the particle = 4.6 x 10⁻⁶N

<em>Substitute these values into equation (i) as follows;</em>

=> 4.6 x 10⁻⁶ = 2.3 x 10⁻⁸ x E

=> E = 4.6 x 10⁻⁶ ÷ 2.3 x 10⁻⁸

=> E = 2 x 10² N/C

Since the value is positive, the electric field is directed upwards.

Therefore, the electric field at that point is +2 x 10² N/C

<em>(b) If a charge of q = -1.1 x 10⁻⁸ is placed there, the force (F) acting is calculated as follows;</em>

<em>Substitute Q = q into equation (i) as follows;</em>

F = q x E

<em>Substitute the value of q and E = 2 x 10² N/C into the equation above as follows;</em>

F = -1.1 x 10⁻⁸ x 2 x 10²

F = -2.2 x 10⁻⁶ N

F = -2.2μN

Since the value of the force, F, is negative, it means it is directed downwards.

Therefore the force on the charge is -2.2μN

3 0

4 years ago