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EastWind [94]
3 years ago
14

With an average acceleration of −1.3 m/s2, how long will it take a cyclist to bring a bicycle with an initial speed of 13.0 m/s

to a complete stop?
Answer in units of s.
Physics
1 answer:
nevsk [136]3 years ago
3 0
13.0÷1.69
= seconds for the cyclist to completely stop

first solve -1.3 to the second power or -1.3 squared by multiplying -1.3 by itself
((-1.3)×(-1.3 )=1.69)
then divide 13 by 1.69
we are dividing because we are solving a deceleration equation and by multiplying 2 negatives the sum will always equal a positive and if we multiplied 13 by 1.69 we would turn this into an accelleration equation.

so divide 13 by 1.69
(13÷1.69=7.6923076923)
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Can somebody help please !!!!
Nataly [62]

Vector A is of magnitude 12 m and it makes an angle of 37 degree with Y axis

So here we can say that

\frac{A_x}{A} = sin37

A_x = A sin37

A_x = 12 sin37

A_x = 7.22 m

Similarly we have

\frac{A_y}{A} = cos37

A_y = A cos37

A_y = 12 cos37

A_y = 9.58 m

So here we have

A_y = 9.58 m, A_x = 7.22 m

option A is correct

3 0
3 years ago
If there is a break at any point in a series circuit the current will
Tomtit [17]

Answer:

not work

Explanation:

in a series circuit, everything meaning the electrons are flowing on one path, therefore, it wouldn continue to work.

8 0
3 years ago
The Doppler effect is the apparent change in frequency due to the movement of the source and the observer. This means an object
Alekssandra [29.7K]

Answer:

Doppler effect changes the wavelength of the light emitted, depending upon whether source is moving away or coming towards the observer(detector).

Explanation:

Doppler effect in light is actually a relativistic effect but somewhat similar to the one which happens in sound waves.

When the source is moving away from the detector, the wavelength of the light emitted from the source appears to be increased as seen by the detector, as a result the frequency decreases(we know that frequency of light= speed of light/wavelength of light. Here speed of light is constant and frequency of light is inversely proportional to its wavelength)

Due to this decrease in frequency the light emitted from the source appears more red, since red color is on low frequency side in the electromagnetic spectrum.

Similarly for the source moving towards the detector, the wavelength appears to be decreased, thereby resulting in increase in frequency and the source appears blue. The shift in frequency is known as doppler shift.

The shift in frequency when the source is moving away is known as redshift and the one where the source is moving towards detector is known as blueshift

4 0
4 years ago
The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 25
Ivanshal [37]

Answer:

a) v(2\,s) = 5.9\,\frac{m}{s}, v(4\,s) = -13.7\,\frac{m}{s}, b) t = 2.602\,s, c) h(2.602\,s) = 35.176\,m, d) t = 5.281\,s, e) v(2.602\,s) = -26.254\,\frac{m}{s}

Explanation:

a) The velocity function is determined by deriving the position function in time:

v(t) = 25.5-9.8\cdot t

Velocities after 2 seconds and 4 seconds are, respectively:

v(2\,s) = 5.9\,\frac{m}{s}

v(4\,s) = -13.7\,\frac{m}{s}

b) The maximum height is reached when velocity is equal to zero:

25.5-9.8\cdot t = 0

The time when the projectile reaches the maximum height:

t = 2.602\,s

c) The maximum height is:

h (2.602\,s) = 2 + 25.5\cdot (2.602\,s)-4.9\cdot (2.602\,s)^{2}

h(2.602\,s) = 35.176\,m

d) The projectile hits the ground when height is equal to zero:

-4.9\cdot t^{2}+25.5\cdot t + 2 =0

The roots of the second order polynomial are presented below:

t_{1} \approx 5.281\,s

t_{2} \approx -0.077\,s

The first one is the only reasonable solution in physical terms.

t = 5.281\,s

e) The velocity of the projectile when it hits the ground is:

v(2.602\,s) = 25.5-9.8\cdot (5.281\,s)

v(2.602\,s) = -26.254\,\frac{m}{s}

4 0
3 years ago
A compact car has a maximum acceleration of 2.0 m/s2 when it carries only the driver and has a total mass of 1100 kg . you may w
nadya68 [22]

According to Newton`s  law. Force exerted by car,

F = m a = 1100 kg \times 2 m/s^2 = 2200 \ N

After adding an additional 400 kg of mass, the force will be same therefore the acceleration

F = 2200 \ N = (1100 \ kg + 400 \ kg)  a \\\\ a = \frac{2200 \ N}{1500 \ kg} = 1.47 \ m/s^2

Thus, the acceleration after adding the masses is 1.47 \ m/s^2.

4 0
3 years ago
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