Answer:
Explanation:
answer C looks good
there isn't really a "centrifugal " force. :/ when we are pushed "back" in a car seat.. it's not because there is a force pushing us backwards... but a force pushing us forwards.. just like when turning too, a force pushes us into the corner, not a force pushing out of the corner. :)
A 1000 kg truck moving at 2.0 m/s runs into a concrete wall. It takes 0.5 s for the truck to completely stop. What is the magnitude of force exerted on the truck during the collision?
Answer:
Momentum is given by
p
=
m
v
. Impulse is the change of momentum,
I
=
Δ
p
and is also equal to force times time:
I
=
F
t
. Rearranging,
F
=
I
t
=
Δ
p
t
=
0
−
20
,
000
5
=
−
4000
N
.
Explanation:
Momentum before the collision is
p
=
m
v
=
2000
⋅
10
=
20
,
000
k
g
m
s
−
1
.
Assuming the truck comes to a complete halt, the momentum after the collision is
0
k
g
m
s
−
1
.
The change in momentum,
Δ
p
, is initial minus final
→
0
−
20
,
000
=
−
20
,
000
This is called the impulse:
I
=
Δ
p
. Impulse is also equal (check the units) to force times time:
I
=
F
t
.
We can rearrange this expression to make
F
the subject:
F
=
I
t
=
Δ
p
t
=
−
20
,
000
5
=
−
4000
N
The negative sign just means the force acting is in the opposite direction to the initial momentum.
(This will be the average force acting during the collision: collisions are chaotic so the force is unlikely to be constant.)
What x rays travel at the speed of
If an object is placed at distance of 16cm from a plane mirror, How far would it be from it's image?
Explanation:
A plane mirror always creates an image with the same distance to the mirror as the object, only in the other direction. So both of them have a distance of 10cm, one is 10cm to the left, one 10cm to the right, thus their mutual distance is 20cm
Two automobiles, each of mass 1000 kg, are moving at the same speed, 20 m/s, when they collide and stick together. In what direction and at what speed does the wreckage move (a) if one car was driving north and one south (b) if one car was driving north and one east?
A. The wreckage after collision is moving at the speed 18 m/s to the south.
B. The wreckage after collision is moving at the speed 9.0 m/s to the north.
C. The wreckage after collision is moving at the speed 9.0 m/s to the south.
D. The wreckage after collision is moving at the speed 18 m/s to the north.
E. The wreckage after collision is motionless.
Answer:
The reckage after collision is motionless (E)
Explanation:
The first law of thermodynamics states that energy is neither created nor destroyed but is converted from one form to another.
The kind of collision described in the question above is known as a perfectly inelastic collision, and in this type of collision, the maximum kinetic energy is lost because the objects moving in opposite directions have a resultant momentum that is equal, but in opposite directions hence they cancel each other out.
The calculation is as follows:
m₁v₁ + m₂v₂
where:
m₁ = m₂ = 1000kg
v₁ = 20 m/s
v₂ = -20 m/s ( in the opposite vector direction)
∴ resultant momentum = (1000 × 20) + (1000 × -20)
= 20000 - 20000 = 0
∴ The reckage after collision is motionless
Answer:
The wreckage after collision is moving at the speed 18 m/s to the south.
Explanation:
4. Name three examples of "concentrated" forms of energy.
Answer:
Nuclear power plant.
Gas stove.
Dam.
Gas pump.
Geothermal heat pump.
Power lines.
Solar panels.
Windmills.
Explanation:
Hope this helps :))
Answer:
gasoline,solar panels,geothermal heat pump,windmills
Explanation:
When you are driving on the freeway and following the car in front of you, how close is too close? Let's do an estimation.
1. Pick a car model (preferably the one you drive, but can also be any car of your dream), and find its stopping distance at highway speeds (you can usually find this type of data online).
2. Assuming that the car in front of you suddenly does a hard brake. For simplicity, assume that its braking performance is about the same as yours. Then also assume a reasonable amount of reaction time on your part (the time delay between seeing the brake lights lit up and applying your own brake). In order for you not to run into the car your are following, what's the closest distance you need to keep between the two cars?
3. Redo the same calculation if the vehicle in front of you is a typical big-rig truck. Find its braking data online.
4. There is a rule of thumb which says that you must stay one car length behind the car in front of you for every 10 mi/h of driving speed. From your calculation, does this rule make sense?
Answer:
1) v= 90km/h d = 70 m, 2) x₁ = v t_r, x₁ = 6.25 m, 3) x₁=6.25 no change
4) x = 22 m
Explanation:
1) for the first part, you are asked to find the minimum safety distance with the vehicle in front
The internet is searched for the stopping distance for two typical speeds on the highway
v (km/ h) v (m/s) d (m)
90 25 70
100 27.78 84
the safe distance is this distance plus the distance traveled during the person's reaction time, which can be calculated with infirm movement
v = x / t_r
x₁ = v t_r
the average reaction time is t_r = 0.25s for a visual stimulus and t_r 0.17 for an auditory stimulus
therefore the safe distance is
x_total = x₁ + d
2) The distance is the sum of the distance traveled in the reaction
x₁ = v t_r
for v = 90 km / h
x₁ = 25 0.25
x₁ = 6.25 m
for v = 100 km / h
x₁ = 27.78 0.25
x₁ = 6.95 m
the total distance is
x_total = x₁ + d
for v = 90 km / h
x_total = 25 0.25 + 70
x_total = 76.25 m
this is the distance until the cars stop and do not collide
3) the stopping distance of a truck is
v = 90 km / h d = 100 m
in this case we see that the braking distance is much higher,
the safe distance is given by the distance traveled during the reaction, as the truck brakes slower than the car this distance does not change
4) let's analyze the empirical rule: maintain the length of a car for each increase in speed of v = 10 m / h = 4.47 m / s
for the car case at v = 90km / h = 25 m / s
according to this rule we must this to
x = 25 / 4.47 = 5.6 cars
each modern car is about 4 m long so the distance is
x = 22 m
we see that this distance is much greater than the reaction distance so it does not make much sense
A physics student sits in a chair. The chair pushes up on the student's body. Identify the other force of the interaction force pair.
Answer:
The other force is the weight of the student.
Explanation:
With respect to Newton's third law of motion, for the student to sit and balance on the chair, there must be two equal and opposite forces involved. The student applies his/ her weight on the chair which acts downwards, while the chair applies an equal but opposite force to the weight of the student.
The force applied by the chair on the student's body is counter balanced by the student's weight. Note that, if the weight of the student is greater than the opposing force from the chair, the chair would collapse.