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3 years ago

A velocity vs time graph what is the acceleration of the object

Physics

2 answers:

tigry1 [53]3 years ago

7 0

Answer:

If the object is moving with an acceleration of +4 m/s/s (i.e., changing its velocity by 4 m/s per second), then the slope of the line will be +4 m/s/s.

Explanation:

Dmitry [639]3 years ago

6 0

Answer:

The graph is missing, but ill try to answer this anyway.

The acceleration is the rate of change of the velocity; this means that if you know the velocity as a function of time; v(t), you can obtain the acceleration a(t) by differentiate v(t) with respect to time; this is:

a(t) = dv(t)/dt

Them if you have a graph of velocity versus time, you can find the equation of v(t) and then find the acceleration.

For example, if your graph of velocity vs time is a linear relation, then the acceleration will be the slope of the linear relation.

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A typical human consumes 2500 Kcal of energy during a day. This is the equivalent to 10,450,000 J! Say you decided to run stairs

defon

Answer:

2940.1 joules would you burn in climbing stairs all day.

Explanation:

Work = W = F $\times$ d

going up stairs would be against force of gravity

W = mgh

where h is the height

the question is not complete because we need speed or distance

h = v $\times$ t

so assuming 1 step per second

h = 86,400 steps $\times$ 7inchs/step $\times$ 0.0254 m/inch

h = 15362 m

so from this

W = 800 N $\times$ 15362

= 12289600 J

that means YOU need 12289600 J to walk 1 step per second all day

divide that by 4180 J /Kcal

Kcal = $\frac{W}{(J/Kcal)}$

= $\frac{12289600}{4180}$

= 2940.1 Kcal

if you ran faster you would use more energy 2 steps per second would mean 5880 Kcal.

8 0

4 years ago

A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.

labwork [276]

Answer:

9.4 m/s

Explanation:

According to the work-energy theorem, the work done by external forces on a system is equal to the change in kinetic energy of the system.

Therefore we can write:

$W=K_f -K_i$

where in this case:

W = -36,733 J is the work done by the parachute (negative because it is opposite to the motion)

$K_i = 66,120 J$ is the initial kinetic energy of the car

$K_f$ is the final kinetic energy

Solving,

$K_f = K_i + W=66,120+(-36,733)=29387 J$

The final kinetic energy of the car can be written as

$K_f = \frac{1}{2}mv^2$

where

m = 661 kg is its mass

v is its final speed

Solving for v,

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s$

4 0

3 years ago

A 3.0kg mass tied to a string

dem82 [27]

Answer:

$\boxed{\sf Tension \ in \ the \ string \ (T) = 3 \ kN}$

Given:

Mass (m) = 3.0 kg

Uniform speed (v) = 20 m/s

Length of string (r) = 40 cm = 0.4 m

To Find:

Tension in the string (T)

Explanation:

Tension (T) is the string will be equal to centripetal force ( $\sf F_c$ ).

$\boxed{ \bold{ T = F_c = \frac{m {v}^{2} }{r} }}$

Substituting value of m, v & r in the equation:

$\sf \implies T = \frac{3 \times {20}^{2} }{0.4} \\ \\ \sf \implies T = \frac{3 \times 400}{0.4} \\ \\ \sf \implies T =3 \times 1000 \\ \\ \sf \implies T =3000 \: N \\ \\ \sf \implies T =3 \: kN$