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
alex41 [277]
3 years ago
9

A state patrol officer saw a car start from rest at a highway​ on-ramp. She radioed ahead to another officer 20 mi along the hig

hway. When the car reached the location of the second officer 19 min​ later, it was clocked going 60​ mi/hr. The driver of the car was given a ticket for exceeding the 60​ mi/hr speed limit. Why can the officer conclude that the driver exceeded the speed​ limit?
Physics
1 answer:
Vikentia [17]3 years ago
5 0

Answer:

We know from the basic speed distance relation that

Speed=\frac{Distance}{Time}

Since the car started from rest and it covered the distance between the 2 officer's in 19 minutes we have speed of the car

Speed=\frac{Distance}{Time}\\\\Speed=\frac{20}{\frac{19}{60}}=63.16mph

Which clearly exceeds the limit of 60\frac{mi}{hr}

You might be interested in
A sprinter accelerates from rest to 10.0m/s in 1.35s.
ANEK [815]

Answer:

a) Acceleration = 7.41 m/s^2

b) Distance ran = 6.75 m

Explanation:

For Acceleration use the formula...

a = (v - u) ÷ t....where v=10 and u=0....and t=1.35

a = (10 - 0)÷ 1.35

;acceleration = 7.41m/s^2

For the distance moved in 1.35s

;use the...s = ut + 1/2at^2....where s is the distance moved

;s = (0)(1.35) + 0.5(7.41 × 1.35^2)

Distance moved = 6.75m

3 0
4 years ago
Read 2 more answers
When two point charges are a distance dd part, the electric force that each one feels from the other has magnitude F.F . In orde
Harman [31]

Answer:

In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. r' = r/2.

Explanation :

The electric force between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between charges. It is given by :

F=\dfrac{kq_1q_2}{r^2}

r is the separation between charges  

F\propto \dfrac{1}{r^2}

r=\sqrt{\dfrac{1}{F}}

If F'= 2F

r'=\dfrac{1}{\sqrt{2F} }

In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. r'=\dfrac{1}{\sqrt{2F} }. Hence, this is the required solution.                                                                                    

6 0
3 years ago
When a force is applied to an object, no work is being preformed on the object unless the object____.
Temka [501]
No work is being performed on the object unless the object is moving, as work done = force x distance
8 0
3 years ago
A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a
EastWind [94]

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2} (1)

Where:

p_{S,1}, p_{S,2} - Initial and final momemtums of the small object, measured in kilogram-meters per second.

p_{B,1}, p_{B,2} - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that p_{S,1} = 7\,\frac{kg\cdot m}{s}, p_{B,1} = 0\,\frac{kg\cdot m}{s} and p_{S, 2} = 4\,\frac{kg\cdot m}{s}, then the final momentum of the big object is:

7\,\frac{kg\cdot m}{s} + 0\,\frac{kg\cdot m}{s} = 4\,\frac{kg\cdot m}{s}+p_{B,2}

p_{B,2} = 3\,\frac{kg\cdot m}{s}

The magnitude of the large object's momentum change is:

p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}-0\,\frac{kg\cdot m}{s}

p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

4 0
3 years ago
A cannonball is fired at a 45.0° angle and an initial velocity of 670 m/s. Assume no air resistance. How long until the cannonba
Naddik [55]

Answer:

96.7 s

Explanation:

Time of flight in projectile can be calculated thus:

T = 2 × u × sin ϴ/ g

Where;

T = time of flight (s)

u = initial velocity (m/s)

ϴ = Angle of projectile (°)

g = acceleration due to gravity (9.8m/s²)

Based on the provided information; u = 670m/s, ϴ = 45°

Hence, using T = 2.u.sin ϴ/ g

T = 2 × 670 × sin 45° ÷ 9.8

T = 1340 × 0.7071 ÷ 9.8

T = 947.52 ÷ 9.8

T = 96.68

T = 96.7s

5 0
3 years ago
Other questions:
  • What would happen if earth gravity suddenly disappeared?
    9·2 answers
  • A circular loop of flexible iron wire has an initial circumference of 167 cm, but its circumference is decreasing at a constant
    9·1 answer
  • Is silicone a better conductor then silver
    7·1 answer
  • URGENT PLS ANSWER When sunlight strikes a solar panel, what form of energy is it transformed to?
    8·2 answers
  • What’s the answer to this?
    6·1 answer
  • find the angle between two forces of magnitude 27N and 30N if the magnitude of the resultant of the two force is 8N​
    7·1 answer
  • We reduce friction in machines? why<br>​
    11·2 answers
  • Y=-3x +4 find the domain Nd range
    14·1 answer
  • The conservation of momentum is most closely related to Newton's?
    14·1 answer
  • Which statement about electromagnets is correct? A. As the number of turns in the wire increases, the strength of the electromag
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!