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AveGali [126]
2 years ago
7

What quantity is represented by a unit called a newton (N)?

Physics
1 answer:
zavuch27 [327]2 years ago
5 0

\textsf {C. Force}

\textsf {The quantity which is represented by a unit called Newton (N)}\\\textsf {is Force.}

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The free-fall acceleration on Mars is 3.7 m/s^2. What length of pendulum has a period of 1.0 s on Earth? What length of pendulum
NemiM [27]

Answer:

Explanation:

Given

Free fall acceleration on mars g_{m}=3.7\ m/s^2

Time Period of pendulum on earth T=1\ s

Time period of Pendulum is given by

T=2\pi \sqrt{\frac{L}{g}}

for earth

1=2\pi\cdot \sqrt{\frac{L}{9.8}}

L=\frac{9.8}{(2\pi )^2}

L=0.498\ m

(b)For same time period on mars length is given by

L'=\frac{g_m}{(2\pi )^2}

L'=\frac{3.7}{39.48}

L'=0.0936\ m

L'=9.36\ cm                            

3 0
3 years ago
A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1
Archy [21]

Answer:

a)  y= 3.5 10³ m, b)   t = 64 s

Explanation:

a) For this exercise we use the vertical launch kinematics equation

Stage 1

          y₁ = y₀ + v₀ t + ½ a t²

          y₁ = 0 + 0 + ½ a₁ t²

Let's calculate

         y₁ = ½ 16 10²

         y₁ = 800 m

At the end of this stage it has a speed

        v₁ = vo + a₁ t₁

        v₁ = 0 + 16 10

        v₁ = 160 m / s

Stage 2

        y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²

        y₂ = 800 + 150 5 + ½ 11 5²

        y₂ = 1092.5 m

Speed ​​is

        v₂ = v₁ + a₂ t

        v₂ = 160 + 11 5

        v₂ = 215 m / s

The rocket continues to follow until the speed reaches zero (v₃ = 0)

         v₃² = v₂² - 2 g y₃

         0 = v₂² - 2g y₃

         y₃ = v₂² / 2g

         y₃ = 215²/2 9.8

         y₃ = 2358.4 m

The total height is

          y = y₃ + y₂

          y = 2358.4 + 1092.5

          y = 3450.9 m

           y= 3.5 10³ m

b) Flight time is the time to go up plus the time to go down

Let's look for the time of stage 3

          v₃ = v₂ - g t₃

          v₃ = 0

          t₃ = v₂ / g

          t₃ = 215 / 9.8

          t₃ = 21.94 s

The time to climb is

          t_{s} = t₁ + t₂ + t₃

          t_{s} = 10+ 5+ 21.94

          t_{s} = 36.94 s

The time to descend from the maximum height is

          y = v₀ t - ½ g t²

When it starts to slow down it's zero

         y = - ½ g t_{b}²

         t_{b}  = √-2y / g

         

        t_{b} = √(- 2 (-3450.9) /9.8)

        t_{b} = 26.54 s

Flight time is the rise time plus the descent date

        t = t_{s} + t_{b}

        t = 36.94 + 26.54

        t =63.84 s

        t = 64 s

3 0
3 years ago
A bumper cart has a mass of 200 kg and has a protective bumper around it that behaves like a spring. The spring constant is 5000
34kurt
Part A:
For this part we’re assuming all the kinetic energy of the moving bumper car is converted into elastic potential energy in the spring since the car is brought to rest. Therefore you can find the total kinetic energy to get your answer:

KE = ½ mv^2
KE = ½ (200)(8)^2
KE = 6400 J

Part B:
Now you can use Hooke’s law to find the force:

F = kx
F = (5000)(0.2)
F = 1000 N
4 0
3 years ago
Net force of 8.0 N acts on an 18 kg body for one minute. Determine the impulse due to the force.
boyakko [2]

Answer:

p = FΔt = 8.0 N(60 s) = 480 N•s

Explanation:

not asked for, but in that time a frictionless 18 kg mass on a horizontal surface will have change velocity by 480/18 = 26.7 m/s.

An impulse results in a change of momentum.

3 0
3 years ago
If you had 8 balls and 7 of them were a certain weight, and 1 of them was heavier, how could you find the heaviest ball. All the
dybincka [34]

Answer:

There are two method of comparing the balls 1) using a balance  2) by  only 2 weighings.

Explanation:

There are two method of comparing the balls 1) using a balance  2) by  only 2 weighings.

Make the following groups - --- (1,2,3),(4,5,6),(7,8)

Step 1. compare the Weigh (1,2,3) and (4,5,6)

there are 2 possible outcomes:

1---both the group are of same weight. and named as (Case A)

2--- one of the group is heavier than other and named as  (Case B)

Step 2. Let examine both case

In Case A --in this case, now compare the weight of 7th and 8th ball. By this you have recognize the heavier ball by 2 weighing method.

In Case B -- considered the heaviest group (assume group (1,2,3) is heavy), from this group take randomly two ball and compare the their weight. out of these two ball, one  is heavy else the third ball is.

7 0
3 years ago
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