Find the kinetic energy of a ball with a mass of 0.06 kg moving at 50 m/s

What is the proper battery cable connection when jumping two automotive batteries? (a) negative to negative / positive to positi

Vladimir [108]

The proper battery cable connection when jumping two automotive batteries is : (a) negative to negative / positive to positive.

Connect the red (positive) cable from the car with the bad battery to the red (positive) on the good battery. 

Then connect the black (negative) from the good battery to a grounding point on the other car which should be tightened and metal should be clean.

Once the car with bad battery has started, the removal of the cable should be in the opposite order. The Red (positive) which was the the First Cable to go on should be the last cable to be taken off.

3 0

3 years ago

The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1

Vaselesa [24]

Answer:

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Explanation:

Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:

$F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2})$ (1)

Where:

$F$ - Explosive force, measured in newtons.

$\Delta s$ - Barrel length, measured in meters.

$m$ - Mass of the shell, measured in kilograms.

$v_{o}$ , $v_{f}$ - Initial and final speeds of the shell, measured in meters per second.

If we know that $m = 1250\,kg$ , $v_{o} = 0\,\frac{m}{s}$ , $v_{f} = 750\,\frac{m}{s}$ and $\Delta s = 15\,m$ , then the explosive force experienced by the shell inside the barrel is:

$F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}$

$F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}$

$F = 23437500\,N$

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

6 0

2 years ago

What is the acceleration of the the object during the first 4 seconds?

AVprozaik [17]

Answer:

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

Explanation:

What this formula is telling us is that if we know the acceleration of an object, and the ... we can plug in our acceleration of 12.5 m/s2 for a, and 4 seconds for t.

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

6 0

3 years ago

A car initially traveling at 17.1 mph comes to rest in 9.7s what was its acceleration in this time?

ra1l [238]

Answer:

$a=-.78m/s^2$

Explanation:

Δ $v=at$

Δ $v$ is the difference in velocity before and after a given time.
$a$ is the acceleration of the object during this time.
$t$ is time

$(v_f-v_i)=at$ is another way to write this equation.

The Δ symbol represents "the difference between the initial and final values of a magnitude or vector", so Δ $v=(v_f-v_i)$

$v_f-v_i=at\\\frac{at}{t}=\frac{v_f-v_i}{t}\\a=\frac{v_f-v_i}{t}$

I rearranged this equation to solve for $a$ , but this is a step that you don't need to take, it's just good to get in the habit of doing this.
Plug in the given values. Note that our final velocity is $0$ , because the car travels until at rest.

$a=\frac{v_f-v_i}{t}\\a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}$

Our initial velocity is in mph, something not in standard units, so if not changed, you will get an incorrect answer. What you need to do is cancel out the units your prior value had using division and multiplication, and at the same time multiply and divide the correct numbers and units into your equation. Or look up a converter.

$a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}\\a=\frac{0m/s-7.6m/s}{9.7s} \\a=\frac{-7.6m/s}{9.7s}$

if you converted correctly, your answer for $v_f$ will be ≅ $7.6m/s$ .
Now divide. Notice that the units for acceleration are $m/s^2$ or meters per second, per second.

$a=\frac{-7.6m/s}{9.7s}\\a=-.78m/s^2$

Our final answer is negative because the car is slowing down. Do not square this answer as the square symbol only applies to the units, not the magnitude.

4 0

3 years ago

Read 2 more answers

When connecting a step down transformer what does the h1 symbol denote?

densk [106]

Answer: Primary coil

A Step-down transformer is an electrical device which is used to decrease the voltage using magnetic induction. The windings of the primary coil are more than the windings of the secondary coils. h1, h2 notations are used to refer to primary coils. On the other hand, x1, x2 notations are used to refer to the secondary coils of the transformer. In case of step-up transformer, the windings of primary coil are less than the secondary in order to increase the voltage.

8 0

3 years ago