All categories

3 years ago

A cyclist maintains a constant velocity of

1 answer:

3 0

The answer is 509 m.

Let point B be 253 m from point A. Let point C be 64 s away from point B.
Let d1 be the displacements from point A to point B and d2 and be the displacements from point B to point C
Step 1. Calculate the displacement from the point B after 64 s.
Step 2. Calculate the displacement from the point A by summing up two distances (d1 and d2).

Step 1.
v = d2/t
v = 4 m/s
d2 = ?
t = 64s
____
4 = d2/64
d2 = 64 * 4 = 256 m

Step 2:
d = d1 + d2
d1 = 253 m
d2 = 256 m
d = 253 + 256 = 509m

You might be interested in

A neutron at rest decays (breaks up) to a proton and an electron. Energy is released in the decay and appears as kinetic energy

SashulF [63]

Answer:

$5.444\times 10^{-4}$

Explanation:

The momentum of the neutron before and after the decay is the same since there's no external force.

$P_{sys}=const\\\\P=mv\\\\K=0.5mv^2$

#The neutron is initially at rest, so after the decay:

$P_A+P_B=0\\\\P_A=-P_B$

#After decay, the proton has +ve direction with a velocity $v_A$ while the electron moves in a negative direction with a velocity $v_B$

Therefore:

$P_A=m_Av_A, P_B=m_Bv_B\\\\\therefore m_Av_A,=m_Bv_B$

Let the energy released during the decay be Q:

$Q=K_{tot}=K_A+K_B\\\\Q=K_A+0.5m_Bv_B^2\\\\Q=K_A+0.5m_B(\frac{m_A}{m_B})^2v_A^2\\\\\ But \ K_A=0.5m_Av_A^2\\\\\therefore Q=K_A+\frac{m_A}{m_B}K_A=K_A(1+\frac{m_A}{m_B})\\\\=Q=\frac{m_A+m_B}{m_B}K_A\\\\m_A=1836m_B\\\\\frac{K_A}{Q}=\frac{m_B}{1836m_B+m_B}=\frac{1}{1837}\\\\\frac{K_A}{Q}=5.444\times10^{-4}$

Hence,Kp/Ktot is 5.444x10^(-4)

4 1

3 years ago

If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder

kakasveta [241]

Question:

A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?

Answer:

Time for the race will be t = 9.26 s

Explanation:

Given data:

As the sprinter starts the race so initial velocity = v₁ = 0

Distance = s₁ = 20 m

Acceleration = a = 4.20 ms⁻²

Distance = s₂ = 100 m

We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.

Using 3rd equation of motion

(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)

v₂ = 12.96 ms⁻¹

Time for 20 m distance = t₁ = (v₂ - v ₁)/a

t₁ = 12.96/4.2 = 3.09 s

He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be

Time for 100 m distance = t₂ = s₂/v₂

t₂ = 80/12.96 = 6.17 s

Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s

T = 9.26 s

5 0

3 years ago

Please help me with question B.

Step2247 [10]

It accelerates in the y component (bc of gravity) AND the x-component (b/c of the friction force).

5 0

3 years ago

Name of compounds<br>a.MgC12<br>b.Mg3 (PO)2<br><br>c.Ca(HCO)2

Vlad1618 [11]

Answer:

a. Magnesium dichloride

b. Magnesium phosphate

c. Calcium hydrogen carbonate

5 0

3 years ago

2. An electric motor transfers 2003 in 10.0 s. If it then transfers the same amount of energy in 8.0 s, has its

Marizza181 [45]

Answer:

decreased

Explanation:

because it transfered from 10.0s

to 8.0s

5 0

2 years ago