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Minchanka [31]
3 years ago
6

A car traveling at 38 m/s starts to decelerate steadily. It comes to a complete stop in 7 seconds. What is its acceleration?

Physics
1 answer:
Katena32 [7]3 years ago
8 0

a  = 38 m/s /  7s

a  =  5.43 ms⁻²

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A car drives on a circular road of radius R. The distance driven by the car is given by = + [where a and b are constants, and t
pishuonlain [190]

Answer:

The question is incomplete, the complete question is "A car drives on a circular road of radius R. The distance driven by the car is given by d(t)= at^3+bt [where a and b are constants, and t in seconds will give d in meters]. In terms of a, b, and R, and when t = 3 seconds, find an expression for the magnitudes of (i) the tangential acceleration aTAN, and (ii) the radial acceleration aRAD3"

answers:

a.18a m/s^{2}

b. a_{rad}=\frac{(27a +b)^{2}}{R}

Explanation:

First let state the mathematical expression for the tangential acceleration and the radial acceleration.

a. tangential acceleration is express as

a_{tan}=\frac{d|v|}{dt} \\

since the distance is expressed as

d=at^{3}+bt

the derivative is the velocity, hence

V(t)=\frac{dd(t)}{dt}\\V(t)=3at^{2}+b\\

hence when we take the drivative of the velocity we arrive at

a_{tan}=\frac{dv(t)}{dt}\\ a_{tan}=6at\\t=3 \\we have \\a_{tan}=18a m/s^2

b. the expression for the radial acceleration is expressed as

a_{rad}=\frac{v^{2}}{r}\\a_{rad}=\frac{(3at^{2} +b)^{2}}{R}\\t=3\\a_{rad}=\frac{(27a +b)^{2}}{R}

4 0
2 years ago
Kindly answer the question about Work and Power. Image is attached below.
natita [175]

Answer:

89600 Watts.

Explanation:

From the question given above, the following data were obtained:

Mass (m) of speedboat = 2800 Kg

Initial velocity (u) = 0 m/s

Final velocity (v) = 16 m/s

Time (t) = 8 s

Power (P) expended by the speedboat =?

Next, we shall determine the acceleration of the speedboat. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 16 m/s

Time (t) = 8 s

Acceleration (a) =?

v = u + at

16 = 0 + (a × 8)

16 = 0 + 8a

16 = 8a

Divide both side by 8

a = 16/8

a = 2 m/s²

The acceleration of the speedboat is 2 m/s²

Next, we shall determine the force exerted by the speedboat. This can be obtained as follow:

Mass (m) of speedboat = 2800 Kg

Acceleration (a) of speedboat = 2 m/s²

Force (F) of speedboat =?

F = ma

F = 2800 × 2

F = 5600 N

Finally, we shall determine the expended by the speedboat at maximum speed. This can be obtained as follow:

Force (F) of speedboat = 5600 N

Maximum velocity (v) = 16 m/s

Power (P) expended by the speedboat =?

P = F × v

P = 5600 × 16

P = 89600 Watts

Thus, the power expended by the speedboat is 89600 Watts.

4 0
2 years ago
A 9-volt battery is used to power two lightbulbs in a series circuit. The first light bulb has a
butalik [34]

Explanation:

Total Resistance: 3+2 = 5 ohms

V=9V

Power = V²/R

=9²/5

=81/5

=16.2 W

6 0
2 years ago
The Hubble Space Telescope is stabilized to within an angle of about 2-millionths of a degree by means of a series of gyroscopes
Likurg_2 [28]

Answer:

The answer to the question is;

The required torque that it would take to cause the gyroscopes to precess through an angle of 1.0×10−6 degree during a 5.0-hour exposure of a galaxy is 2.44 ×10⁻¹² N·m  

Explanation:

To solve the question we first resolve the units of the given quantities as follows

The gyroscopes spin at 19,200 rpm that is 19,200 revolutions per minute

1 revolution = 2π rad and

1 minute = 60 seconds

Therefore 19,200 revolutions per minute = 2π×19,200÷60 rad/s

= 2010.619 rad/s

The angle of precess is given as 1.0×10⁻⁶ °. We convert the angle to radians as follows

360 ° = 2π radians

1 ° = \frac{\pi }{180} radians and

1.0×10⁻⁶ ° =  \frac{\pi }{180} radians × 1.0×10⁻⁶ ° = 1.745×10⁻⁸ rad

To find the torque we note that the torque is given by

Precession angular speed × The moment of inertia × angular velocity

The precession angular speed is given by \frac{Precession. Angle}{time}

The precession angle was determined in rad as 1.745×10⁻⁸ rad

The precession time is 5 hours which is equal to 5×60×60 = 18000 s

Therefore the precession velocity = \frac{1.745*10^{-8} rad}{18000 s} =  9.696×10⁻¹³ rad/s

The moment of inertia is given by

Formula for the moment of inertia of a thin walled cylinder I = m·r²

Where:

r = Radius of the gyroscope = Diameter/2 = 5.0 cm/2 = 2.5 cm = ‪0.025‬ m

m = Mass of each gyroscopes = 2.0 kg

Therefore I = m·r² = 2.0 kg × (0.025‬ m)² = 0.00125 kg·m²

Torque, τ = Ω·I·ω

Where:

Ω = Precession velocity = 9.696×10⁻¹³ rad/s

I = Moment of inertia = 0.00125 kg·m²

ω = Angular speed = 2010.619 rad/s

τ = 9.696×10⁻¹³ rad/s × 0.00125 kg·m² × 2010.619 rad/s =

2.44 ×10⁻¹² kg·m²/ s² =   2.44 ×10⁻¹² N·m  

The required torque is 2.44 ×10⁻¹² N·m.

7 0
3 years ago
Which property of X-rays allows them to enter our bodies and create pictures of our bones and other internal structures?
Degger [83]

Answer:

(A) Their high energy

Explanation:

The higher the energy of the x-ray the more "likely" they will pass through.

The x-ray is electromagnetic energy caused by energy changes in an electron.

X-ray are energetic and have enough energy to break up molecules. That is what reveals the bad or damaged cells in our body.

7 0
3 years ago
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