Someone help please !!

A train takes ms eaddy 200 miles north and 150 miles south. The total trip took 3.5 hours what was the average velocity of the t

Verizon [17]

The whole point of this problem is to check how well you understand
the definitions of a few important quantities, like velocity, speed, distance,
displacement etc.

Before we begin, I just want to mention that 'MPG' is not a unit of either
velocity or speed, but I think I know what you mean.

-- For some reason, Ms. Eaddy rode 100 miles north on the train, then
stayed aboard while the train turned around and took her 150 miles south.
The total distance she rode was (100 + 150) = 250 miles. But she ended up
50 miles south of where she began.

-- Displacement for the whole trip = distance and direction from the start point
to the finish point.

Displacement = 50 miles south

-- Average velocity = (displacement) / (time)

50 miles south / 3.5 hours = <u>14.29 miles per hour south</u>

3 0

3 years ago

Homes with multiple bathrooms have outlets connected in series with a red reset button in the system

dybincka [34]

Homes with multiple bathrooms have outlets connected in series with a red reset button in the system. The reason for this circuit design is to prevent electrocution.

<h3>What is red reset button?</h3>

The water heater's reset switch provided on the upper thermostat of an electric water heater. If the button is on, the switch would have tripped and the reset is required.

To prevent any kind of injury or killing of a person by electric shock is called as the electrocution. This can be prevented only by outlets connected in series with a red reset button.

Thus, the reason for this circuit design is to prevent electrocution.

Learn more about red reset button.

brainly.com/question/7643711

#SPJ1

8 0

2 years ago

A school bus has a mass (including the driver and passengers) of 1.64 times 10^4 kg and is driving north at a speed of 15.2 km/h

Butoxors [25]

Answer:

Explanation:

Given

mass of bus along with travelers travelling in North direction is $m_1=1.6\times 10^4 kg$

speed of bus towards North $v_1=15.2 km/h\approx 4.22\ m/s$

mass of bus travelling in South direction is $m_2=1.578\times 10^4 kg$

speed of bus $v_2=12.2 km/h\approx 3.38\ m/s$

mass of each Passenger in south moving bus $m_0=64.8 kg$

Momentum of North moving bus

$P_1=m_1\times v_1$

$P_1=1.6\times 10^4\times 4.22$

$P_1=6.768\times 10^4 kg-m/s$

Momentum with south moving bus

$P_2=m_2\times v_2+n\cdot m_0\times v_2$

$P_2=(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38$

For total momentum to be towards south

$P_2-P_1$ should be greater than 0

thus for least value of n

$P_2=P_1$

$(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38=6.768\times 10^4$

$1.578\times 10^4+n\cdot 64.8=2.0023\times 10^4$

$n=\frac{4243.6686}{64.8}=65.48\approx 66$

5 0

3 years ago

A truck initially traveling at a speed of 22 m/s increases at a constant rate of 2.4 m/s^2 for 3.2s. What is the total distance

FinnZ [79.3K]

Answer:

82.7 m

Explanation:

u= 22m/s

a= 2.4 m/s^2.

t= 3.2 secs

Therefore the distance travelled can be calculated as follows

S= ut + 1/2at^2

= 22 × 3.2 + 1/2 × 2.4 × 3.2^2

= 70.4 + 1/2×24.58

= 70.4 + 12.29

= 82.7 m

Hence the distance travelled by the truck is 82.7 m

6 0

3 years ago

Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the

nydimaria [60]

Answer: $F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

$F_1=\frac{kq(-q)}{(L\sqrt{2})^2}$

where $L\sqrt{2}=$ Distance between the two charges

$F_1=-\frac{kq^2}{2L^2}$

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

$F_2=\frac{kq(-q)}{(L)^2}$

The magnitude of force by both the charge is same but at an angle of $90^{\circ}$

thus combination of two forces at 2 and 3 will be

$F'=\sqrt{2}\frac{kq^2}{2L^2}$

Now it will add with force due to 1 charge

Thus net force will be

$F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

6 0

3 years ago