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3 years ago

9

estimate the force that a cyclist travelling at 6m/s needs to apply to the brakes to bring the bicycle to a stop in a distance o

f 15m. Ignore air resistance & friction.

1 answer:

RSB [31]3 years ago

4 0

v1 = 6m/s

v2 = 0

∆v = v1 - v2 = 6m\s

s = t * v = 15m

t = s\v1 = 15(m) \ 6(m\s) = 2.5s

a = ∆v\t = 6(m\s) \ 2.5s = 2.4m\s2

a = F\m = 2.4m\s2

F = a * m = 2.4m\s2 * ?kg

I can't tell you this because I don't know the mass of this cyclist

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A projectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. The two protons then mo

makkiz [27]

Answer:

(a) The speed of the target proton after the collision is: $V_{2f} =433(m/s)$ , and (b) the speed of the projectile proton after the collision is: $v_{1f}=250(m/s)$ .

Explanation:

We need to apply at the system the conservation of the linear momentum on both directions x and y, and we get for the x axle: $m_{1} v_{1i} =m_{1} v_{1f}Cos\beta _{1} +m_{2} v_{2f}Cos\beta _{2}$ , and y axle: $0=m_{1} v_{1f}Sin\beta _{1}+m_{2} v_{2f}Sin\beta _{2}$ . Now replacing the value given as: $v_{1i}=500(m/s)$ , $\beta_{1}=+60^{o}$ for the projectile proton and according to the problem $\beta_{1}and\beta_{2}$ are perpendicular so $\beta_{2}=-30^{o}$ , and assuming that $m_{1}=m_{2}$ , we get for x axle: $500=v_{1f}Cos\beta _{1}+ v_{2f}Cos\beta _{2}$ and y axle: $0=v_{1f}Sin\beta _{1}+v_{2f}Sin\beta _{2}$ , then solving for $v_{2f}$ , we get: $v_{2f}=-v_{1f}\frac{Sin\beta_{1}}{Sin\beta_{2}}= \sqrt{3}v_{1f}$ and replacing at the first equation we get: $500=\frac{1}{2} v_{1f} +\frac{\sqrt{3}}{2} *\sqrt{3}*v_{1f}$ , now solving for $v_{1f}$ , we can find the speed of the projectile proton after the collision as: $v_{1f}=250(m/s)$ and $v_{2f}=\sqrt{3}*v_{1f}=433(m/s)$ , that is the speed of the target proton after the collision.

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3 years ago

1. You serve a volleyball with a mass of 2.1kg. The ball leaves your

Darina [25.2K]

The kinetic energy is 945 joules.

Kinetic energy is the energy that an object has as a result of motion. It is defined as the effort required to accelerate a mass-determined body from rest to the indicated velocity.

The speed of an object or particle, which is a scalar quantity, is the size of the change in its location over time or the size of the change in its position per unit of time.

The mass of the volleyball is 2.1 kg.

The speed of the ball when the ball leaves the hand is 30 m/s.

m = 2.1 kg

v = 30 m/s

The kinetic energy of an object is given as:

KE = (1/2 ) × m × v²

KE = (1 / 2) × 2.1 kg × ( 30 m/s)²

KE = (1 / 2) × 2.1 kg × 30 m/s × 30 m/s

KE = 2.1 kg × 15 m/s × 30 m/s

KE = 945 J

Learn more about kinetic energy here:

brainly.com/question/8101588

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1 year ago

A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectivel

faust18 [17]

Answer:

The time is 5.10 sec.

Explanation:

Given that,

Component of velocity are 30 m/s and 40 m/s.

We need to calculate the resultant velocity

Using formula of resultant velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(30)^2+(40)^2}$

$v=50\ m/s$ [

We need to calculate the time

Using equation of motion

$v = u+gt$

$v = 0+gt$

$t=\dfrac{v}{g}$

Put the value into the formula

$t=\dfrac{50}{9.8}$

$t=5.10\ sec$

Hence, The time is 5.10 sec.

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3 years ago

What did the asymptote say to the removable discontinuity worksheet answers?

ra1l [238]

“Don't hand that holier than thou line to me” is what the asymptote said to the removable discontinuity.

The distance between the curve and the line where it approaches zero as they tend to infinity is the line in the asymptote of a curve. This is unusual for modern authors but in some sources the requirement that the curve may not cross the line infinitely often is included.

The point that does not fit the rest of the graph or is undefined is called a removable discontinuity. By filling in a single point, the removable discontinuity can be made connected.

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