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2 years ago

A race car accelerates at a constant rate from 20 m/s to 60 m/s in a time of 2 seconds. What is the car’s acceleration?

Physics

1 answer:

Vlada [557]2 years ago

5 0

$\huge\fbox{Answer ☘}$

Given ,

initial velocity , u = 20 m/s

final velocity , v = 60 m/s

time taken = 2 seconds

Now ,

$\bold{acceleration = \frac{v - u}{t} } \\ \\ = > \bold{ \frac{60 - 20}{2} } \\ \\ = > \bold{\frac{40}{2} } \\ \\ \bold{ => 20 \: ms {}^{ - 2}}$

hope helpful~

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Explain how machines can be useful if the output is always less than the input work

Anni [7]

Because: Some of the work done by the machine is used to overcome the friction created by the use of the machine. ... Work output can never be greater than work input. Machines allow force to be applied over a greater distance, which means that less force will be needed for the same amount of work.

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3 years ago

1) A force of 20 Newton acts on a bar having a cross sectional area of 0.8m^2 and length 10cm.calculate the stress developed in

Elanso [62]

Answer:25N/M^2

Explanation:

Force=20N Area=0.8M^2

Stress=force/area

Stress=20/0.8

Stress=25N/M^2

4 0

3 years ago

The electronics supply company where you work has two different resistors, R1 and R2, in its inventory, and you must measure the

elixir [45]

Answer:

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Explanation:

When connected in series, the equivalence resistance, R(eq) is given as

R(eq) = (R₁ + R₂)

When connected in parallel, the equivalence resistance, R(eq) is given as

[1/R(eq)] = [(1/R₁) + (1/R₂)]

R(eq) = (R₁R₂)/(R₁ + R₂)

The parallel and series combination are connected to a battery of emf 39.0 V with negligible internal resistance. And the power supplied is measured.

But power supplied is given as

P = IV = (V/R) V = (V²/R)

When connected in series, the power supplied is given as

P = 48.0 W,

V = 39.0 V,

R = R(eq) = (R₁ + R₂)

48 = (39²/R)

R = (39²/48)

R = 31.6875 ohms

R = (R₁ + R₂) = 31.6875

(R₁ + R₂) = 31.6875 (eqn 1)

When connected in series, the power supplied is given as

P = 256.0 W,

V = 39.0 V,

R = R(eq) = (R₁R₂)/(R₁ + R₂)

256 = (39²/R)

R = (39²/256)

R = 5.9414 ohms

R = R(eq) = (R₁R₂)/(R₁ + R₂) = 5.9414

(R₁R₂)/(R₁ + R₂) = 5.9414

But, recall eqn 1

(R₁ + R₂) = 31.6875

(R₁R₂)/(R₁ + R₂) = 5.9414

Substituting for (R₁ + R₂)

(R₁R₂)/(R₁ + R₂) = (R₁R₂)/31.6875 = 5.9414

(R₁R₂) = 31.6875 × 5.9414 = 188.2683

R₁ = (188.2683/R₂)

(R₁ + R₂) = 31.6875

Substituting for R₁

(188.2683/R₂) + R₂ = 31.6875

multiply through by R₂

188.2683 + R₂² = 31.6875R₂

R₂² - 31.6875R₂ + 188.2683 = 0

Solving the quadratic equation

R₂ = 23.77 ohms or 7.92 ohms

If R₂ = 23.77 ohms, R₁ = 7.92 ohms

If R₂ = 7.92 ohms, R₁ = 23.77 ohms

Since the question explains that R₁ > R₂

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Hope this Helps!!!

3 0

3 years ago

Please need help fast

iVinArrow [24]

(a) See graph in attachment

The appropriate graph to draw in this part is a graph of velocity vs time.

In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.

Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at

t = 10 s

v = 15 m/s

(b) $1.5 m/s^2$

The average acceleration of the horse can be calculated as:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

In this problem,

u = 0

v = 15 m/s

t = 10 s

Substituting,

$a=\frac{15-0}{10}=1.5 m/s^2$

(c) 75 m

For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance travelled

u is the initial velocity

t is the time interval

a is the acceleration

In this problem,

u = 0

t = 10 s

$a=1.5 m/s^2$

Substituting,

$s=0+\frac{1}{2}(1.5)(10)^2=75 m$

(d) See attached graphs

In a uniformly accelerated motion:

- The distance travelled (x) follows the equation mentioned in part c,

$x=ut+\frac{1}{2}at^2$

So, we see that this has the form of a parabola: therefore, the graph x vs t will represents a parabola.

- The acceleration is constant during the motion, and its value is

$a=1.5 m/s^2$ (calculated in part b)

therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.

6 0

3 years ago

Electromagnetic waves differ from other types of waves because they are (2 points)

saveliy_v [14]

Answer:

b. able to travel through a vacuum.

Explanation:

The most distinguishing factor of an electromagnetic waves is that they are able to travel through a vacuum.

These waves do not require materials in a medium for propagation.

Electromagnetic waves are formed by the propagation of the electric and magnetic fields.
They vibrate at an angle of 90° .
They are unlike like mechanical waves that requires that requires materials in medium for their propagation.

7 0

3 years ago