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ollegr [7]
4 years ago
9

A heavy 2.0Kg ball is moving at 10m/s when it is caught. A light 1.5Kilogram ball is travelling at 20m/s when it caught. Which b

all required the greater impulse? a) The 2.0kg ball because it is more massive. b) The 2.0kg ball because it had more momentum c) The 1.5kg ball because it was travelling faster d) The 1.5 kg ball because it is more momentum e) The question can’t be answered without more information.
Physics
2 answers:
Daniel [21]4 years ago
7 0

To check impulse you will need more information you will require more data as it was diverse with time. In this question, there are two balls with various mass and speed, along these lines they have diverse force. I will expect you are getting some information about the momentum.


The momentum of ball 1 should be:

momentum 1 = mass₁ x velocity₁

momentum 1 = 2kg x 10m/s = 20kg m/s


The momentum of ball 2 should be:

momentum 2= mass₂ x velocity₂

momentum 2= 1.5kg x 20 m/s = 30kg m/s


The answer would be "1.5kg ball because it was travelling faster".

kupik [55]4 years ago
6 0
The answer is d) The 1.5 kg ball because it has more momentum. 
Momentum = Impulse = Mass + Velocity
= 2 * 10 = 20 kg*m/s
= 1.5 * 20 = 30 kg*m/s
hence the 1.5 kg ball has greater momentum and impulse. 

Source: http://forumvu.com/Thread-GSC101-Assignment-1-Solution-Fall-2017
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Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to
stiks02 [169]

Now, there is some information missing to this problem, since generally you will be given a figure to analyze like the one on the attached picture. The whole problem should look something like this:

"Beam AB has a negligible mass and thickness, and supports the 200kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously."

Answer:

\mu_{sB}=0.126

\mu_{sC}=0.168

Explanation:

In order to solve this problem we will need to draw a free body diagram of each of the components of the system (see attached pictures) and analyze each of them. Let's take the free body diagram of the beam, so when analyzing it we get:

Sum of torques:

\sum \tau_{A}=0

N(3m)-W(1.5m)=0

When solving for N we get:

N=\frac{W(1.5m)}{3m}

N=\frac{(1962N)(1.5m)}{3m}

N=981N

Now we can analyze the column. In this case we need to take into account that there will be no P-ycomponent affecting the beam since it's a slider and we'll assume there is no friction between the slider and the column. So when analyzing the column we get the following:

First, the forces in y.

\sum F_{y}=0

-F_{By}+N_{c}=0

F_{By}=N_{c}

Next, the forces in x.

\sum F_{x}=0

-f_{sB}-f_{sC}+P_{x}=0

We can find the x-component of force P like this:

P_{x}=360N(\frac{4}{5})=288N

and finally the torques about C.

\sum \tau_{C}=0

f_{sB}(1.75m)-P_{x}(0.75m)=0

f_{sB}=\frac{288N(0.75m)}{1.75m}

f_{sB}=123.43N

With the static friction force in point B we can find the coefficient of static friction in B:

\mu_{sB}=\frac{f_{sB}}{N}

\mu_{sB}=\frac{123.43N}{981N}

\mu_{sB}=0.126

And now we can find the friction force in C.

f_{sC}=P_{x}-f_{xB}

f_{sC}=288N-123.43N=164.57N

f_{sC}=N_{c}\mu_{sC}

and now we can use this to find static friction coefficient in point C.

\mu_{sC}=\frac{f_{sC}}{N}

\mu_{sC}=\frac{164.57N}{981N}

\mu_{sB}=0.168

3 0
3 years ago
On December 26, 2004, a violent earthquake of magnitude 9.1 occurred off the coast of Sumatra. This quake triggered a huge tsuna
Sonja [21]

Answer:

a) 800 km

b) 9.79° & 12.47°

Explanation:

It is known that the wavelength of any material or substance is given by the relation 入 = v/f

From the question, the time between the crest is given as 1.0 h, therefore, t = 1.0 h

Also, remember that frequency, f can be represented by

f = 1/t, and thus,

f = 1/1.0 h = 1.0 h^-1. This is our frequency, and would be used in the primary stated formula

入 = v/f, from the question, the speed is already stated as 800 km/h, so we apply directly

入 = 800 / 1

入 = 800 km. Therefore, the wavelength of the tsunami is 800 km

B,

the smallest angle between the Southern tip of Africa is

Sinθ = 800 / 4500

Sinθ = 0.17

θ = Sin^-1 0.17

θ = 9.79°

The smallest angle between the Southern end of Australia is

Sinθ = 800 / 3700

Sinθ = 0.216

θ = Sin^-1 0.216

θ = 12.47°

4 0
4 years ago
Calculate the approximate force on a square meter (1.00 m3) of sail, given the horizontal velocity of the wind is 6.00 m/s paral
KiRa [710]

Answer:

The force exerted on square meter of cubic sail is F = 15.3 N  

Explanation:

Given:-

- The face area of cubic sail, A = 1 m^2

- The velocity at frontal face, v1 = 6.0 m/s

- The velocity at back face, v2 = 3.5 m/s

- The density of air. ρ = 1.29 kg/m^3

Find:-

Calculate the approximate force on a square meter (1.00 m3) of sail

Solution:-

- We will apply the Bernoulli's equation to the flow of air around the cubic sail. Assuming that elevation changes are negligible. The constant elevation Bernoulli's equation is:

                           P1 + ρ*v1^2 / 2 = P2 + ρ*v2^2 / 2  

- The force (F) applied by any fluid is given by:

                           F = ( P2 - P1 )*A

- Re-arranging bernoulli's expression:

                          P2 - P1 = ρ/ 2 [ v2^2 - v1^2 ]

- Multiple the equation by area A:

                          A*[P2 - P1] = A*ρ/ 2 [ v1^2 - v2^2 ]

                          F = A*ρ/ 2 [ v1^2 - v2^2 ]

- Plug in the values:

                         F = (1)*(1.29/2)*[ 6^2 - 3.5^2 ]

                         F = 15.3 N  

7 0
4 years ago
Write a sentence to describe how an image is formed on retina<br>when looking at distant objects.​
beks73 [17]

Answer:

An image is formed on the retina with light rays converging most at the cornea and upon entering and exiting the lens. Rays from the top and bottom of the object are traced and produce an inverted real image on the retina. The distance to the object is drawn smaller than scale

7 0
3 years ago
2. Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m. The charges are +7.0μC, -8.0 μC
Scorpion4ik [409]

Answer:

0.53 N, 25.6°

Explanation:

side of triangle, a = 1.2 m

q = 7 μC

q1 = - 8 μC

q2 = - 6 μC

Let F1 be the force between q and q1

By using the coulomb's law

F_{1}=\frac{Kq_{1}q}{a^{2}}

F_{1}=\frac{9\times 10^{9}\times 7\times 10^{-6}\times 8\times 10^{-6}}{1.2^{2}}

F1 = 0.35 N

Let F2 be the force between q and q2

By using the coulomb's law

F_{2}=\frac{Kq_{2}q}{a^{2}}

F_{2}=\frac{9\times 10^{9}\times 7\times 10^{-6}\times 6\times 10^{-6}}{1.2^{2}}

F2 = 0.26 N

Write the forces in the vector form

\overrightarrow{F_{1}}=0.35\widehat{i}

\overrightarrow{F_{2}}=0.26\left ( Cos60 \widehat{i}+Sin60\widehat{j}\right )

\overrightarrow{F_{2}}=0.13 \widehat{i}+0.23\widehat{j}

Net force

\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}

\overrightarrow{F}=0.48 \widehat{i}+0.23\widehat{j}

Magnitude of the force

F=\sqrt{0.48^{2}+0.23^{2}}

F = 0.53 N

Direction of force with x axis

tan\theta =\frac{0.23}{0.48}

θ = 25.6°

7 0
3 years ago
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