Energy that a microwave use to heat food?

Issac and Blaise decide to race. They both start at the same position at the same time. Issac runs at 2m/s but decides to take a

FromTheMoon [43]

Let the Blaise runs for time "t" to complete the race

so the total distance he moved is given by

$d_1 = 1* t$

Now Issac runs for time t = "t - 2*60"

because it took rest for 2 minutes

$d_2 = 2*(t - 120)$

now it is given that Blaise wins by 10 m distance

$d_1 - d_2 = 10$

$1* t - 2*(t - 120) = 10$

$t - 2t + 240 = 10$

$t = 230 s$

now the distance moved by Blaise is given by

$d_1 = 1*230 = 230 m$

6 0

3 years ago

How many seconds will it take for a the International Space Station to travel 450 km at a rate of 100 m/s?

SVEN [57.7K]

Time = (distance) / (speed)

Time = (450 km) / (100 m/s)

Time = (450,000 m) / (100 m/s)

Time = 4500 seconds (that's 75 minutes)

Note:

This is about HALF the speed of the passenger jet you fly in when you go to visit Grandma for Christmas.

If the International Space Station flew at this speed, it would immediately go ker-PLUNK into the ocean.

The speed of the International Space Station in its orbit is more like 3,100 m/s, not 100 m/s.

8 0

3 years ago

Calculate the kinetic energy of a moving 4 kg object traveling at a velocity of 3 m/a

Alina [70]

Answer:

Explanation:KE = 18 J

7 0

3 years ago

JOSEPH JOGS FROM END A TO OTHER END B OF A straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back

zzz [600]

(a) The average speed from A to B would be 1.76 metre per second and the average velocity from A to B would also be 1.76 metre per second

(b) The average speed from A to C would be 1.73 metre per second and the average velocity from A to C would be 0.87 metre per second

3 0

3 years ago

A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio

natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

$\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}$

For the maximum velocity of the ball at the top of the vertical circular motion,

$v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}$

where g is the acceleration due to gravity,

Substituting the known values,

$\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}$

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

8 0

1 year ago