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marysya [2.9K]
3 years ago
10

In the example of jumping off a chair, what is the impulse that will stop your fall?

Physics
1 answer:
liq [111]3 years ago
5 0

Answer:

You will reach both your arms out to break your fall and save your head.

Explanation:

It common sense you don't want your head injured. Do you?

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A 1500 kg car traveling due east at 20<br> m/s slows to a stop in 5.0 seconds. What is the impulse?
lbvjy [14]

Answer:

Hans-Georg Gadamer (1900-2002) was an influential German philosopher of the twentieth century, inspiring a variety of scholastic disciplines from aesthetics to theology. In suggesting understanding was interpretation and vice versa, Gadamer identifies language acting as the medium for understanding. Gadamer’s philosophy of hermeneutics has major implications for education and formal schooling because Hermeneutics help to know the knowledge a student has prior to the lesson. This helps in the dialogue about a subject matter and therefore, the philosophy of Hermeneutics when applied in classroom helps the teachers pass information easily and effectively, hence, the learners capture the whole content of a topic.

Explanation:

5 0
2 years ago
Read 2 more answers
A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
tatuchka [14]

Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

so:

I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

7 0
3 years ago
Which factors affect the strength of the electric force between two objects
Margaret [11]
<span>-- the product of the net charges on the objects;. -- the distance between the centers of their net charges. (Pretty much identical to the formula for gravitational force)</span>
8 0
3 years ago
Read 2 more answers
A baseball has a mass of 0.145 kilograms, and a bowling ball has a mass of 6.8 kilograms. What is the gravitational force betwee
Olegator [25]
Quite low

Gravitation force = \frac{6.67* 10^{-11}*0.145*6.8 }{ 0.5^{2} } = 26.3 * 10^{-11} = 2.63 *10^{-10} N
6 0
2 years ago
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A 26.4 g silver ring (cp = 234 J/kg·°C) is heated to a temperature of 66.2°C and then placed in a calorimeter containing 4.94 ✕
Slav-nsk [51]

Answer:

The final temperature of the mixture = 64.834 °C.

Explanation:

Heat lost by the silver ring = heat gained by the water + heat transferred to the surrounding.

c₁m₁(t₁-t₃) = c₂m₂(t₃-t₂) + Q..............Equation 1

Where c₁ = specific heat capacity of the silver copper, m₁ = mass of the silver copper, t₁ = initial temperature of the silver copper, t₃ = final temperature of the mixture. c₂ = specific heat capacity of water, t₂ = initial temperature of water, m₂ = mass of water, Q = energy transferred to the surrounding.

making t₃ the subject of the equation,

t₃ = [c₁m₁t₁+c₂m₂t₂-Q]/(c₁m₁+c₂m₂)........................ Equation 2

Given: c₁ = 234 J/kg.°C, m₁ = 26.4 g, t₁ = 66.2 °C, c₂ = 4200 J/K.°C, m₂ = 4.92×10⁻² kg, t₂ = 24.0 °C, Q = 0.136 J.

Substituting into equation 2

t₃ = [(234×26.4×66.2)+(4200×0.0492×24)-0.136]/[(234×26.4)+(4200×0.0492)]

t₃ = (408957.12+4959.36-0.136)/(6177.6+206.64)

t₃ = (413916.48-0.136)/6384.24

t₃ = 413916.34/6384.24

Thus the final temperature of the mixture = 64.834 °C.

6 0
3 years ago
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