Answer:
The correct answer is - 1240 newtons; 278.763 lbs.
Explanation:
What is the weight of a 25 kg object on Earth with an acceleration due to gravity of 9.8m/s/s?
2.45 n
24.5 n
245 n
2450 n
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
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:
You are riding in the passenger seat of a car as it goes around a tight turn. You slide across the seat to the passenger side door. Which statement below properly describes what is happening? a. You are exerting a centripetal force on the door. b. The door and seat are exerting a centripetal force on you that balances the centrifugal force of the turn. c. The car seat exerts a centripetal force on you, but not enough to keep you in place, so the door exerts the rest. d. The centrifugal force is greater than the force of friction between you and the seat, so you slide outward.
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. :)
19. In a turtle race, a 20 kg turtle moves with a velocity of 0.1 m/s. What is the kinetic energy of the turtle?
A. 20 J
B. 0.2 J
C. 0.1 J
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
Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for slinging supplies to affected areas, similar to the electromagnetic lift catapults used to launch planes from aircraft carriers. This catapult is located at a fixed point 400 meters away and 50 meters below the target site. The catapult is capable of launching the payload at 67 meters per second and an initial launch angle of 50 degrees. Using your knowledge of kinematics equations, determine whether this would be sufficient to deliver the payload to the drop site.
Answer:
Please see below as the answer is self-explanatory.
Explanation:
We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).Both movements are independent each other, due to they are perpendicular.In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:[tex]v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)[/tex]
Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:[tex]t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s (2)[/tex]
In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:[tex]\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (3)[/tex]
In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:[tex]v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)[/tex]
Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:[tex]\Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)[/tex]
Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.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