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erik [133]
3 years ago
9

Please help!!

Physics
2 answers:
Sveta_85 [38]3 years ago
6 0

Answer:

If you push hard, the block's bottom will move out from under its center of gravity and the block will tip over. The taller the block, the easier it is to push the bottom out from under the center of gravity. Keeping a car's center of gravity low is important to avoiding disaster during turns. But if you don't keep the car's center gravity low it will be inversely proportional and the acceleration can be damaged with force, and its mass could get disproportionate.

Explanation:

I'm sorry i tried my best. :/

baherus [9]3 years ago
4 0

Answer:

Mass is the correct answer.

Explanation:

Many drivers report a more positive handling response and a definite improvement when reducing unsprung mass. You want to keep unsprung weight to as little as possible. This minimizes the momentum and energies that your suspension has to counter. In effect, it can make your shocks more sensitive.

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At what time was the person at a position of 0m?
Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line  equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

  = (0 x 1) + 1/2 (0 x 1 x 1)

  = (0) + 1/2 (0)

  = 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

 = 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1)  =  10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

8 0
3 years ago
What provides the force on the person on the passenger seat?
Gennadij [26K]
There's the acceleration of the car that provides a force and the normal force of the seat cushion which pushes upwards against the passenger
4 0
3 years ago
If an object has a mass of 1417g and it is moved 47 meters in 90 seconds, how much power was used?
astraxan [27]

Mass of the object is given as

m = 1417 g = 1.417 kg

now the speed of object is given as

v = \frac{d}{t}

here we know that

d = 47 m

t = 90 s

now we will have

v = \frac{47}{90} = 0.52 m/s

now we will have kinetic energy of the object as

KE = \frac{1}{2}mv^2

KE = \frac{1}{2}(1.417)(0.52)^2

KE = 0.19 J

now the power is defined as rate of energy

so here we can find power as

P = \frac{KE}{t}

P = \frac{0.19}{90} = 2.14\times 10^{-3} W

so above is the power used for the object

3 0
3 years ago
When a piano tuner strikes both the A on the piano and a 440 Hz tuning fork, he hears a beat every 2 seconds. The frequency of t
AysviL [449]

Answer:

The  frequency is  f = 439.5 \  Hz        

Explanation:

From the question we are told that

   The frequency of the  tuning fork is  f_t  =  440 \ Hz

   The beat period is  T  =  2 \  s

Generally the beat frequency is mathematically represented as

       f_b  =  \frac{1}{T}

       f_b  =  \frac{1}{2}

      f_b  = 0.5 \  Hz

The  beat frequency is also represented mathematically as

     f_b  =  f_t  -  f

Where  f is the frequency of the piano

 So

       f =  440 -  0.5  

       f = 439.5 \  Hz        

3 0
3 years ago
According to newton’s second law of motion, the acceleration of an object equals the net force acting on the object divided by t
Ugo [173]
C mass I am pretty sure
3 0
3 years ago
Read 2 more answers
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