A baseball pitcher throws a ball across home plate. The ball travels 13.40m in 0.357s Determine the

The maximum speed limit on interstate 10 is 75 miles per hour. how many meters per second is this

Dvinal [7]

Answer:

Explanation:

Given the maximum speed limit on interstate 10 as 75 miles per hour, to get the speed in meter per seconds, we need to convert the given speed to meter per seconds.

Using the conversion 1 mile = 1609.34m and 1 hour = 3600 seconds

75 miles perhour = 75miles/1 hour

75miles/1 hour (in m/s) = 75miles*1609.34m* 1 hour/1mile * 1 hour * 3600s *

= 75 *1609.34m* 1 /1 * 1 * 3600s

= 120,700.5m/3600s

= 33.53m/s

<em>Hence the maximum speed limit on interstate 10 in metre per seconds is 33.53m/s</em>

8 0

3 years ago

So velocity of a balloon? Use V = 2KE over v. A hot air balloon with a mass of 100 kg moves across the sky with 3200 J of kineti

kicyunya [14]

Answer:

700

Explanation:

just put it in kokkokokokokokoook

7 0

3 years ago

3. Identify the structures of the skin by the labels A through G below

ioda

I need help to I have a big test and I need help so bad

4 0

3 years ago

Convert 1nanosecond in to its SI init

SOVA2 [1]

<em>Convert 1nanosecond in to its SI init</em>

<em>In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.</em>

<em>In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.So to convert nanosecond into second, just multiply the nanosecond with 0.000000001 or (10^-9)</em>

8 0

3 years ago

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

8 0

3 years ago