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bonufazy [111]
3 years ago
7

In anticipation of a long 10 upgrade, a bus driver accelerates at a constant rate of 5 ft/s2 while still on a level section of a

highway. Knowing that the speed of the bus was 80 mph as it begins to go upthe hill and that the driver doesnot change the setting on his throttle or shift gears, determine the distance traveled (in miles) by the bus upthe hill when its speed decreased to 50mph.Problem
Physics
1 answer:
fgiga [73]3 years ago
3 0

Answer:

S = 20903.4 ft

Explanation:

First we will determine the acceleration of the bus while it is moving upward

The equilibrium  equation would be

F - W sin\theta = ma'\\ma - W sin\theta = ma'\\m = \frac{W}{g} \\\frac{W}{g}*a - W sin\theta = \frac{W}{g} * a'\\\frac{a}{g} - sin\theta = \frac{a'}{g}\\\frac{x}{y} \frac{5}{32.2} - sin 10 = \frac{a'}{32.2} \\\a' = -0.59

Let the displacement be S_0

As per newton's third law of motion

v^2 -u^2 = 2as\\u = 80 \frac{mi}{h} =   117.33\frac{ft}{s}\\v = 50 \frac{mi}{h} =   73.33\frac{ft}{s}\\73.33 ^ 2 - 117.33^2 =  2 * (-0.59) * (S-S_0)\\\\S_0 = 0\\S = 20903.4

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Hydrogen does not contain any neutrons

7 0
3 years ago
Read 2 more answers
Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block
Oksana_A [137]

Answer:

av=0.333m/s, U=3.3466J

b.

v_{A2}=-1.333m/s,\\ v_{B2}=0.667m/s

Explanation:

a. let m_A be the mass of block A, andm_B=10.0kg be the mass of block B. The initial velocity of A,\rightarrow v_A_1=2.0m/s

-The initial momentum =Final momentum since there's no external net forces.

pA_1+pB_1=pA_2+pB_2\\\\P=mv\\\\\therefore m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}

Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):

v_A_1-v_B_1=v_{B2}-v_{A2}

-Applying the conservation of momentum. The blocks have the same velocity after collision:

v_{B2}=v_{A2}=v_2\\\\2\times 2+10\times 0=2v_2+10v_2\\\\v_2=0.3333m/s

#Total Mechanical energy before and after the elastic collision is equal:

K_1+U_{el,1}=K_2+U_{el,2}\\\\#Springs \ in \ equilibrium \ before \ collision\\\\U_{el,2}=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\\\\U_{el,2}=0.5\times 2\times 2^2-0.5(2+10)(0.333)^2\\\\U_{el,2}=3.3466J

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0

We plug these values in the equation:

m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}

2\times2+10\times0=2v_A_2+10v_B_2\\\\2=v_A_2+5v_B_2\\\\#Eqtn 2:\\v_A_1-v_B_1=v_{B2}-v_{A2}\\\\2-0=v_{B2}-v_{A2}\\\\2=v_{B2}-v_{A2}\\\\#Solve \ to \ eliminate \ v_{A2}\\\\6v_{B2}=2.0\\\\v_{B2}==0.667m/s\\\\#Substitute \ to \ get \ v_{A2}\\\\v_{A2}=\frac{4}{6}-2=1.333m/s

7 0
4 years ago
10. Calculate the kinetic energy of a running back that has a mass of 80 kg and
EastWind [94]

Answer:

The answer is

<h2>2560 J</h2>

Explanation:

The kinetic energy of an object given it's mass and velocity can be found by using the formula

KE =  \frac{1}{2} m {v}^{2}

where

m is the mass

v is the velocity

From the question

m = 80 kg

v = 8 m/s

The kinetic energy is

KE =  \frac{1}{2}  \times 80 \times  {8}^{2}  \\  = 40 \times 64

We have the final answer as

<h3>2560 J</h3>

Hope this helps you

5 0
3 years ago
The the figure shows a famous roller coaster ride. You can ignore friction. If the roller coaster leaves Point Q from rest, what
ch4aika [34]

Answer:

22 m/s

Explanation:

PEf +KEf =PE0 +KE0 →PE0 −PEf =KEf

−mgΔy= 1 mv2 →v= −2gΔy = −2(9.8 m/s2)(−25 m)=22 m/s

3 0
3 years ago
A charged particle moving along the +x-axis enters a uniform magnetic field pointing along the +z-axis. A uniform electric field
Brums [2.3K]

Answer:

the electric field direction should be in positive y axis

Explanation:

Lets assume that charge on particle is positive and it isequal to +q

First calculate the magnetic force on it

F_B = q V\times B =qVBsin\theta

for direction

use Right Hand Rule which will give the direction and by using his the direction will come towards negative y axis.

As given in the question that charge particle does not change their velocity so we need to apply electric field in such a way that electric force direction should be opposite to the magnectic field.

and magnitude should be same as magnectic force and also direcion of electric force depend on the direction of elecric field when charge is positive because electric force F_E = qE

Hence the electric field direction should be in positive y axis

3 0
4 years ago
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