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3 years ago

What is indicated by the slope of an acceleration vs. time graph?

Physics

1 answer:

Luba_88 [7]3 years ago

4 0

Answer:

Option A is tge correct answer.

Explanation:

The reason for above answer is the slope of acceleration vs time graph shows the velocity as when the acceleration gets positive , the velocity gets increased and when the acceleration gets negative tgen velocity also decreases.

hope it helps..

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What type of energy results from the burning of wood or gasoline?

Sonja [21]

Answer:

Chemical energy

Explanation:

Chemical energy is energy stored in the bonds of atoms and molecules. Batteries, biomass, petroleum, natural gas, and coal are examples of chemical energy. Chemical energy is converted to thermal energy when people burn wood in a fireplace or burn gasoline in a car's engine.

7 0

2 years ago

The driver of a car going 86.0 km/h suddenly sees the lights of a barrier 44.0 m ahead. It takes the driver 0.75 s to apply the

Lynna [10]

Part a

Answer: NO

We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.

Using the second equation of motion:

$s=ut+0.5at^2$

where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.

It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, $a=-10.0 m/s^2$

$\Rightarrow s=23.9 m/s \times 0.75s+0.5\times (-10.0 m/s^2)\times (0.75 s)^2=15.11 m$

Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.

Part b

Answer: 29.6 m/s

The maximum distance that car can travel is $s=44.0 m$

The acceleration is same, $a=-10.0 m/s^2$

The final velocity, v=0

Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

$v^2-u^2=2as$

$0-u^2=-2 \time 10.0m/s^2 \times 44.0 m\Rightarrow u=\sqrt{880 m^2/s^2}=29.6 m/s$

Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.

7 0

3 years ago

On August 10, 1972, a large meteorite skipped across the atmosphere above the western United States and western Canada, much lik

Anon25 [30]

a) $4.62\cdot 10^{14} J$

b) 0.110 megatons

c) 8.46 bombs

Explanation:

The energy lost by the meteorite is equal to the difference between its final kinetic energy and its initial kinetic energy:

$\Delta K=K_f-K_i$

Which can be rewritten as:

$\Delta K=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

where:

$m=3.2\cdot 10^6 kg$ is the mass of the meteorite

$v=0$ is the final speed of the meteorite

$u=17 km/s = 17,000 m/s$ is the initial speed of the meteorite

Substituting the values into the equation, we found the loss in energy of the meteorite:

$\Delta K=0-\frac{1}{2}(3.2\cdot 10^6)(17000)^2=-4.62\cdot 10^{14} J$

So, the energy lost by the meteorite is $4.62\cdot 10^{14} J$

The energy equivalent to 1 megaton of TNT is

$E_{TNT}=4.2\cdot 10^{15} J$

Here the energy lost by the meteorite is

$E=4.62\cdot 10^{14} J$

Therefore, in order to write the energy lost by the meteorite as a multiple of the energy of 1 megaton of TNT, we have to divide the energy lost by the meteorite by the energy equivalent to 1 TNT; we find:

$\frac{E}{E_{TNT}}=\frac{4.62\cdot 10^{14}}{4.2\cdot 10^{15}}=0.110$

So, the energy lost by the meteorite corresponds to 0.110 megatons.

The energy of one atomic bomb explosion in Hiroshima is equal to

$E'=13 kt$ (13 kilotons)

which corresponds to

$E'=0.013 Mt$ (0.013 megatons)

Here the energy of the meteorite is equal to

$E=0.110 Mt$ (0.110 megatons)

Therefore, we can find how many Hiroshima bombs are equivalent to teh meteorite impact by using the following rules of three:

$\frac{1 bomb}{0.013 Mt}=\frac{x bombs}{0.110 Mt}\\x=\frac{1\cdot 0.110}{0.013}=8.46$

So, 8.46 bombs.

5 0

3 years ago

Which of the following can be determined by studying the spectrum of a star?

Mama L [17]

The right answer for the question that is being asked and shown above is that: "all of the above." the following that can be determined by studying the spectrum of a star includes: the chemicals in the atmosphere of the star; the motion of the star; the temperature of the atmosphere of the star.

8 0

3 years ago

Read 2 more answers

Does heavy force work under conditions of weightlessness in such an artificial earth satellite?

Afina-wow [57]

Answer: That is to say, a satellite is an object upon which the only force is gravity. Once launched into orbit, the only force governing the motion of a satellite is the force of gravity. Newton was the first to theorize that a projectile launched with sufficient speed would actually orbit the earth.

Explanation: Because Earth-orbiting objects follow elliptical paths around Earth and not a straight line, forces cannot, by definition, be balanced. Force is directional. It is a push or a pull in a particular direction.

3 0

2 years ago