Ask question

Ask question

All categories

3 years ago

7

An object of mass 3.05 kg, moving with an initial velocity of 5.09 i m/s, collides with and sticks to an object of mass 2.34 kg

with an initial velocity of -3.28 j m/s. find the final velocity of the composite object.

1 answer:

Irina18 [472]3 years ago

3 0

Try using the G.U.E.S.S method. Which is G for givens which you write down all the information it gives you in the problem. U for the unknown like what are you trying to find/what the question asking for. E for equation find the right formula that best fits this problem. S for substitute so plug in the numbers given in the problem into the formula that best fits. The last S is solve which you workout the equation to find what the question is asking for. Hope this helps :)

You might be interested in

Suppose that you measure the length of a spaceship, at rest relative to you, to be 400 m. how long will you measure it to be if

Fantom [35]

Calculate the length of a spaceship as follows:

l = l₀√1 - v²/c²

=(400 m)√1 - (0.75c)2 c²

=264.575m.

The Spaceship Origin Portfolio is an index fund that invests in listed Australian and global equities by market capitalization. Invest in the top 100 Australian and top 100 international companies.

Starships, also known as star cruisers, starships, spacecraft, or simply starships or ships, were vessels designed specifically for interstellar travel between star systems.

For clients in the Spaceship Index portfolio, the situation is a little different. The Spaceship Index portfolio consists of approximately 100 of his ASX-listed companies with the largest market capitalization and approximately 100 global companies with the largest market capitalization.

Learn more about spaceship at

brainly.com/question/28175986

#SPJ4

8 0

2 years ago

What could be the possible answer to the question ?<br><br>thankyou ~

Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

$T_1 = \mathbf{\dfrac{M \cdot g}{2}}$

Which gives;

$M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}$

$T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})} = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}$

F₀ = T₂·sin(37°)

Which gives;

$F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2} \approx \mathbf{0.377 \cdot M \cdot g}$

F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

3 0

2 years ago

Read 2 more answers

A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/

Elena-2011 [213]

First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

$KE_i + PE_i = KE_f + PE_f$

$\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0$

$\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2$

So by solving above equation we will have

$v_f = 84.3 m/s$

now in order to find the momentum we can use

$P = mv$

$P = 1.25 \times 84.3$

$P = 105.4 kg m/s$

6 0

3 years ago

A certain car engine delivers enough force to create 630 N⋅m of torque when the engine is operating at 3200 revolutions per minu

jekas [21]

The appropriate expression for the calculation of power by relating the angular energy in a given time.

In other words the instantaneous power of an angular accelerating body is the torque times the angular velocity

$P=\tau\omega$

Where

$\tau =$ Torque

$\omega =$ Angular speed

Our values are given by

$\tau = 630Nm$

$\omega = 3200rev/min$

The angular velocity must be transformed into radians per second then

$\omega = 3200rev/min (\frac{2\pi rad}{60s})$

$\omega = 335.103rad/s$

Replacing,

$P=(630)(335.103)$

$P = 211.11*10^3W$

$P = 211.1kW$

The average power delivered by the engine at this rotation rate is 211.1kW

5 0

3 years ago

A ball is thrown straight up. At the top of its path its acceleration is

timurjin [86]

C. Acceleration is the rate of change of velocity. So at the top of the path, while the velocity is zero, the CONSTANT GRAVITATIONAL ACCELERATION is about 10 m/s^2 (9.8)

6 0

3 years ago