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4 years ago

A bike accelerates uniformly(from rest to a speed of 7 m/s over a distance of 40 m. Determine

Physics

1 answer:

marshall27 [118]4 years ago

3 0

Answer:

$0.61 m/s^2$

Explanation:

The bike's acceleration can be found by using the following suvat equation:

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

where

v is the final velocity of the bike

u is the initial velocity

a is the acceleration

s is the distance covered

For the bike in the problem,

u = 0

v = 7 m/s

d = 40 m

Solving the equation for a, we find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{7^2-0}{2(40)}=0.61 m/s^2$

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svp [43]

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4 years ago

A car horn emits a frequency of 400 Hz. A car traveling at 20.0 m/s sounds the horn as it approaches a stationary pedestrian. Wh

Temka [501]

Answer:

The observed frequency by the pedestrian is 424 Hz.

Explanation:

Given;

frequency of the source, Fs = 400 Hz

speed of the car as it approaches the stationary observer, Vs = 20 m/s

Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.

The observed frequency is calculated as;

$F_s = F_o [\frac{v}{v_s + v} ] \\\\$

where;

F₀ is the observed frequency

v is the speed of sound in air = 340 m/s

$F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\$

F₀ ≅ 424 Hz.

Therefore, the observed frequency by the pedestrian is 424 Hz.

8 0

3 years ago

The magnitude of electrical force between a pair of charged particles is ____ as much when the particles are moved half as far a

Gnesinka [82]

The magnitude of the electrical force between a pair of charged particles is 4 Times as much when the particles are moved half as far apart.

This can be easily understood by Columb's law,

$F_{new} = \frac{kQ_{1}Q_{2}}{r^{2}}$

which state's that the amount of electrical force experienced by two charged particles is inversely proportional to the square of the distance between them.

∴ $\frac{F_{new} }{F_{old} } = \frac{Distance_{new}^{2} }{Distance_{old}^{2} }$

Now, we know the new distance is half the original distance,

$F_{new} = \frac{kQ_{1}Q_{2}}{\frac{r}{2}^{2} } \\F_{new} = 4\frac{kQ_{1}Q_{2}}{r^{2}}$

$F_{new} = 4F_{old}$

The electrical force of attraction or electrostatic force of attraction between two charged particles refers to the amount of attractive or repulsive force that exists between the two charges. This can be calculated by Columb's Law.

A charged particle in physics is a particle that has an electric charge. It might be an ion, such as a molecule or atom having an excess or shortage of electrons in comparison to protons. The same charge is thought to be shared by an electron, a proton, or another primary particle.

Learn more about electrical force here

brainly.com/question/2526815

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8 0

2 years ago

Two particles A and B start simultaneously from a Point P with velocities 20 m/s and 30 m/s respectively. A and B move with acce

zysi [14]

Answer:

20 m/s

Explanation:

Given

u(A) = 20 m/s
u(B) = 30 m/s
acceleration equal in magnitude but opposite in direction

Solving

Velocity of A at Q = 30 m/s
From, P to Q, Δv(A) = 30 - 20 = +10 m/s
Therefore, velocity of B at Q will be decreased by 10 as it is equal in magnitude but opposite in direction to A
Δv(B) = v(B at Q) - u(B at P)
-10 m/s = v(B at Q) - 30 m/s
v(B at Q) = 30 - 10 = 20 m/s

6 0

3 years ago

Can someone please help

Maru [420]

Answer: D

Explanation: because doing a yoga desk program is physical activity, 10k steps is pysical activity, riding a bike or walking/running is also physical activity. so it should be D, all of the above.

5 0

3 years ago