answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
barxatty [35]
1 year ago
13

Automobile A starts from O and accelerates at the constant rate of 0.75 m/s2. A short time later it is passed by bus B which is

traveling in the opposite direction at a constant speed of 6 m/s. Knowing that bus B passes point O 20 s after automobile A started from there, determine when and where the vehicles passed each other.
Physics
1 answer:
boyakko [2]1 year ago
6 0

Answer:

they meet from point o at distance 50.46 m and time taken is 11.6 seconds

Explanation:

given data

acceleration = 0.75 m/s²

speed B = 6 m/s

time B = 20 s

to find out

when and where the vehicles passed each other

solution

we consider here distance = x , when they meet after o point

and time = t for meet point z

we find first Bus B distance for 20 s ec

distance B = velocity × time

distance B = 6 × 20

distance B = 120 m

so

B take time to meet is calculate by distance formula

distance = velocity × time

120 - x = 6 × t

x = 120 - 6t   .................1

and

distance of A when they meet by distance formula

distance = ut + 1/2 × at²

here u is initial speed = 0 and t is time

x = 0 + 1/2 × 0.75 × t²

x = 0.375 × t²      .............2

so from equation 1 and 2

0.375 × t²  = 120 - 6t

t = 11.6

so time is 11.6 second

and

distance from point o from equation 2

x = 0.375 (11.6)²

x = 50.46

so distance from point o is 50.46 m

You might be interested in
The actions of an employee are not attributable to the employer if the employer has not directly or indirectly encouraged the em
zepelin [54]

Answer:    the answer is d

Explanation: there are not more than 10 violations  within a twelve month period hope this helps

4 0
2 years ago
You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
NemiM [27]

Answer:

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 *10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g= GM /R² = (v²*(R+h)) /(R²)

⇒ with v= the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 *10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * ( 4.48*10^6+ 630000)) / ((4.48*10^6)²)

g = 6.11 m/s²

<u>Step 3:</u> Describe the position of the projectile

Horizontal component: x(t) = v0*t *cos∅

Vertical component: y(t) = v0*t *sin∅ -1/2 gt² ( will be reduced to 0 in time )

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t= v(sin∅)/g = 1.14 s

Horizontal range d = v0²/g *2sin∅cos∅  = v0²/g * sin2∅

Horizontal range d =(13.6²)/6.11 * sin(2*30.8)

Horizontal range d =26.63 m

The horizontal range of the projectile = 26.63 meters

6 0
1 year ago
The air in a 6.00 L tank has a pressure of 2.00 atm. What is the final pressure, in atmospheres, when the air is placed in tanks
ser-zykov [4K]

Explanation:

Given that,

Initial volume of tank, V = 6 L

Initial pressure, P = 2 atm

We need to find the final pressure when the air is placed in tanks that have the following volumes if there is no change in temperature and amount of gas:

(a) V' = 1 L

It is a case of Boyle's law. It says that volume is inversely proportional to the pressure at constant temperature. So,

PV=P'V'\\\\P'=\dfrac{PV}{V'}\\\\P'=\dfrac{6\times 2}{1}\\\\P'=12\ atm

(b) V' = 2500 mL

New pressure becomes :

PV=P'V'\\\\P'=\dfrac{PV}{V'}\\\\P'=\dfrac{6\times 2}{2500\times 10^{-3}}\\\\P'=4.8\ atm

(c) V' = 750 mL

New pressure becomes :

PV=P'V'\\\\P'=\dfrac{PV}{V'}\\\\P'=\dfrac{6\times 2}{750\times 10^{-3}}\\\\P'=16\ atm

(d) V' = 8 L

New pressure becomes :

PV=P'V'\\\\P'=\dfrac{PV}{V'}\\\\P'=\dfrac{6\times 2}{8}\\\\P'=1.5\ atm

Hence, this is the required solution.

3 0
2 years ago
A ball is falling at terminal velocity. Terminal velocity occurs when the ball is in equilibrium and the forces are balanced. Wh
Greeley [361]

Answer:

A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N.

Explanation:

This is because at terminal velocity, the ball stops accelerating and the net force on the ball is zero. For the net force to be zero, equal and opposite forces must act on the ball, so that their resultant force is zero. That is F₁ + F₂ = 0 ⇒ F₁ = -F₂

Since F₁ = 20 N, then F₂ = -F₁ = -20 N

So, if F₁ points upwards since it is positive, then F₂ points downwards since it is negative.

So, a free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N best describes the ball falling at terminal velocity.

4 0
2 years ago
Read 2 more answers
You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof w
Anit [1.1K]

Answer:

v₀ =3.8 m/s

Explanation:

Newton's second law of the box:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Known data

m=2.1 kg  mass of the box

d= 5.4m  length of the roof

θ = 20° angle θ of the roof with respect to the horizontal direction

μk= 0.51 : coefficient of kinetic friction between the box and the roof  

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the box on the roof  and the y-axis in the direction perpendicular to it.

W: Weight of the box  : In vertical direction

N : Normal force : perpendicular to the direction the  roof

fk : Friction force: parallel to the direction to the roof

Calculated of the weight  of the box

W= m*g  =  (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y weight components

Wx= Wsin θ= (20.58)*sin(20)° =7.039 N

Wy= Wcos θ =(20.58)*cos(20)°= 19.34 N

Calculated of the Normal force

∑Fy = m*ay    ay = 0

N-Wy= 0

N=Wy =19.34 N

Calculated of the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx-f = ( 2.1)*a

7.039 - 9.86  = ( 2.1)*a

-2.821 = ( 2.1)*a

a=(-2.821) /( 2.1)

a= -1.34  m/s²

Kinematics of the box

Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:  

d:displacement  = 5.4 m

v₀: initial speed  

vf: final speed  = 0

a : acceleration of the box = -1.34  m/s²

We replace data in the formula (2)

0²=v₀²+2*(-1.34)*(5.4)

2*(1.34)*(5.4)= v₀²

v_{o} =\sqrt{14.472}

v₀ = 3.8 m/s

7 0
2 years ago
Other questions:
  • a truss is made by hinging two uniform, 150-N rafters as shown in Fig. 5-21. They rest on an essentially frictionless floor and
    6·1 answer
  • For what value of the ratio r/a of plate radius to separation between the plates does the electric field at the point x=a/2 on t
    15·1 answer
  • A rock is dropped from the top of a tall building. The rock's displacement in the last second before it hits the ground is 46 %
    14·1 answer
  • Joe pushes down the length of the handle of a 10.9 kg lawn spreader. The handle makes an angle of 45.3 ◦ with the horizontal. Jo
    9·1 answer
  • 2. A pebble is dropped down a well and hits the water 1.5 s later. Using the equations for motion with constant acceleration, de
    14·1 answer
  • Two 7.0cm×7.0cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 v battery.
    9·1 answer
  • Astronauts in the International Space Station must work out every day to counteract the effects of weightlessness. Researchers h
    15·1 answer
  • Sasha is ordered Ampicillin 50mg/kg/day x 48 hours, to be given every 6 hours in 100mls of N/S run over 30 minutes. The tubing h
    12·1 answer
  • Based on the free-body diagram, the net force acting on this wheelbarrow is {blank} N.
    15·2 answers
  • A fundraising company agrees to donate an extra $75 for every $100 the school raises through selling cookies. Part A: What is th
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!