All categories

3 years ago

If you run 12 m/s for 15 minutes, how far will you go?

Physics

2 answers:

worty [1.4K]3 years ago

7 0

(12 meter/second) (60 sec/minute) (15 minutes) =

(12 60 15) (meter second minute / second minute) =

10,800 meters (10.8 km, about 6.7 miles)

Note: This question doesn't represent a real-world situation. Usain Bolt's best time for 100 meters ... the current world record ... is a speed of 10.4 m/s !

vfiekz [6]3 years ago

3 0

10800 m = 10.8 km should be the answer if I am correct

You might be interested in

Help on physics problem (picture above)

natali 33 [55]

The two letters are B and A, in that order.

7 0

3 years ago

An object falls freely for 25 seconds, what is the velocity after the 25 seconds ?

Andrei [34K]

The acceleration of gravity on Earth is 9.8 m/s² downward.
This means that gravity adds 9.8 m/s downward to the speed
of a freely falling object every second.

So after 25 sec, it's falling (25 x 9.8m/s) = 245 m/s faster than
it was falling at the beginning of the 25 seconds.

If it dropped from rest (no speed), then its velocity
after 25 seconds is 245 m/s downward.

5 0

3 years ago

Convert 14 minutes to seconds.<br> (What unit do we want)<br><br> 1. Seconds <br> 2. Minutes

Mnenie [13.5K]

Answer:

Seconds...

Explanation:

4 0

3 years ago

Assume that in 2010 the United States will need 2.0×1012 watts of electric power produced by thousands of 1000 MW power plants.

Alex73 [517]

Answer:

1752.14 tonnes per year.

Explanation:

To solve this exercise it is necessary to apply the concepts related to power consumption and power production.

By conservation of energy we know that:

$\dot{P} = \bar{P}$

Where,

$\dot{P} =$ Production of Power

$\bar{P} =$ Consumption of power

Where the production of power would be,

$\dot{P} = m \dot{E}\eta$

Where,

m = Total mass required

$\dot{E} =$ Energy per Kilogram

$\eta =$ Efficiency

The problem gives us the aforementioned values under a production efficiency of 45%, that is,

$\dot{P} = \bar{P}$

$m \dot{E}\eta = \bar{P}$

Replacing the values we have,

$m(8*10^13)(0.45) = 2*10^{12}$

Solving for m,

$m = \frac{ 2*10^{12}}{(8*10^13)(0.45)}$

$m = 0.0556 \frac{kg}{s}$

We have the mass in kilograms and the time in seconds, we need to transform this to tons per year, then,

$m = 0.556\frac{kg}{s}*(\frac{3.1536*10^7s}{1year})(\frac{1ton}{1000kg})$

$m = 1752.14$ tonnes per year.

8 0

3 years ago

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.5 meters

dem82 [27]

Answer:

The electric and magnetic field are 6.34 N/C and $2.11\times10^{-8}\ T$ .

Explanation:

Given that,

Distance = 4.5 m

Time = 2.84 μs

We need to calculate the acceleration

Using equation of motion

The distance covered by the electron is

$s=ut+\dfrac{1}{2}at^2$

When, the electron at rest

$s = \dfrac{1}{2}at^2$

Where, s = distance

a = acceleration

t = time

Put the value into the formula

$4.5=\dfrac{1}{2}\times a\times(2.84\times10^{-6})^2$

$a=\dfrac{2\times4.5}{(2.84\times10^{-6})^2}$

$a=1.116\times10^{12}\ m/s^2$

We need to calculate the electric field

Using formula of the electric field

$E=\dfrac{F}{q}$

$E=\dfrac{ma}{q}$

Put the value into the formula

$E=\dfrac{9.1\times10^{-31}\times1.116\times10^{12}}{1.6\times10^{-19}}$

$E=6.34\ N/C$

We need to calculate the magnetic field

Using formula of magnetic field

$B = \dfrac{E}{c}$

Put the value into the formula

$B=\dfrac{6.34}{3\times10^{8}}$

$B=2.11\times10^{-8}\ T$

According to the right hand rule,

The direction of magnetic field is outward because the direction of force is upward.

Hence, The electric and magnetic field are 6.34 N/C and $2.11\times10^{-8}\ T$ .

8 0

3 years ago