1answer.
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
kirill [66]
3 years ago
12

A train leaves the train station at noon and travels at a constant speed of vt = 50 mi/hr on a straight track. 2 hr later, a car

leaves the same train station and travels in the same direction at vc = 70 mi/hr on a road next to the train track. How far from the station is the place where the car catches up with the train? x =
Physics
1 answer:
asambeis [7]3 years ago
6 0

Answer:

350 miles

Explanation:

When the car starts 2 hours later, the train would have a head start of

50 * 2 = 100 miles

The speed of the car relative to the train is

70 - 50 = 20 mi/hr

For the car to catch up with the train, it must cover the 100 miles difference at the rate of 20mi/hr. So the time it would need to cover this difference is

100 / 20 = 5 hours

After 5 hours, the car would have traveled a distance of

5 * 70 = 350 miles which is also the distance from the station to where the car catches up

You might be interested in
What is less dense water or oil steel or water helium or air or oil or water
Leni [432]

Answer:

air because their is nothing contained within the air other than all the solutions that you have listed

Explanation:

4 0
3 years ago
Read 2 more answers
Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each ot
Zolol [24]

Answer:

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² =  2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ =  m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁  = 4.427 - 2v₂  ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂  = 8.854

v₂ = 8.854 / 3

v₂  = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁  = - 1.477 m/s

v₁  = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

mgh_{max} = \frac{1}{2}mv_{max}^2

(a) The maximum height achieved by the first ball (v₁  = 1.477 m/s)

mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max}  =  \frac{1}{2g}v_{max}^2\\\\ h_{max}  = \frac{1}{2*9.8}(1.477^2)\\\\ h_{max}  = 0.11 \ m

(b) The maximum height achieved by the second ball (v₂  = 2.95 m/s)

mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max}  =  \frac{1}{2g}v_{max}^2\\\\ h_{max}  = \frac{1}{2*9.8}(2.95^2)\\\\ h_{max}  = 0.44 \ m

6 0
3 years ago
An elevator accelerates upward at 1.2 m/s 2 . the acceleration of gravity is 9.8 m/s 2 . what is the upward force exerted by the
bonufazy [111]
The upward force is 737 N
6 0
3 years ago
Calculate the current through the ammeter (look at photo)
AleksAgata [21]

Answer:

6 amps

Explanation:by Kirchhoff's loop rule the current at any point in the loop must be equal or charge would be building up. The current at the ammeter is equally to the total current through the sun of the paths in parallel which it is in series with

8 0
2 years ago
Suppose a 1500 kg speed boat is traveling with a velocity of 30 m/s what is the speed boats Monument
DanielleElmas [232]

Answer: 45000kgm/s

Explanation:

Given that:

Mass of speed boat = 1500 kg

Velocity of speedboat = 30 m/s

Speed boats Momentum = ?

The Speed boat momentum is the product of its mass and the velocity by which it moves. Momentum is a vector quantity and measured in kgm/s

i.e Momentum = mass x velocity

= 1500 kg x 30 m/s

= 45000kgm/s

Thus, the speed boats momentum is 45000kgm/s

7 0
3 years ago
Other questions:
  • When neither air mass displaces the adjacent one, their boundary is called a(n) ________ front?
    15·1 answer
  • A child rides her bike at a rate of 12.0 km/hr down the street. A squirrel suddenly runs in front of her so she applies the brak
    6·2 answers
  • PLEASE ANSWER QUESTION 25! NOT 24. Brainliest and 25 points!
    8·2 answers
  • The displacement is a straight line from the starting point to the finish in a ____ direction
    9·1 answer
  • A boy runs 400m at an average speed of 4.0m/s he runs the first 200m in 40 s how long does he take to run the second 200m?
    11·1 answer
  • A 0.74 mF capacitor is connected to a standard outlet (rms voltage 82 V, frequency 49 Hz ). Determine the magnitude of the curre
    11·1 answer
  • When you push a child on a swing, you are doing work on the child because _____.
    15·2 answers
  • The ideal mechanical advantage of a pulley system is equal to the?
    7·2 answers
  • Please help me with 1&2
    11·1 answer
  • What does law and theory have common
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!