All categories

3 years ago

With what speed does the can move immediately after the collision? Answer in units of m/s.

Physics

1 answer:

Ratling [72]3 years ago

7 0

Answer:

1.74 m/s

Explanation:

From the question, we are given that the mass of the an object, m1= 2.7 kilogram(kg) and the mass of the can,m(can) is 0.72 Kilogram (kg). The velocity of the mass of an object(m1) , V1 is 1.1 metre per seconds(m/s) and the velocity of the mass of can[m(can)], V(can) is unknown- this is what we are to find.

Therefore, using the formula below, we can calculate the speed of the can, V(can);

===> Mass of object,m1 × velocity of object, V1 = mass of the can[m(can)] × velocity is of the can[V(can)].----------------------------------------------------(1).

Since the question says the collision was elastic, we use the formula below

Slotting in the given values into the equation (1) above, we have;

1/2×M1×V^2(initial velocity of the first object) + 1/2 ×M(can)×V^2(final velocy of the first object)= 1/2 × M1 × V^2 m( initial velocity of the first object).

Therefore, final velocity of the can= 2M1V1/M1+M2.

==> 2×2.7×1.1/ 2.7 + 0.72.

The velocity of the can after collision = 1.74 m/s

You might be interested in

Assume the height of the roller coaster (see fig) is y = 40 m (take the reference point, y = 0, as the bottom of the

PtichkaEL [24]

H⁣⁣⁣⁣ere's l⁣⁣⁣ink t⁣⁣⁣o t⁣⁣⁣he a⁣⁣⁣nswer:

bit. $^{}$ ly/3a8Nt8n

7 0

3 years ago

Calculate the force of gravity between planet X and planet y if both planets are 3.75 X 10^11 m apart, planet X has a mass of 1.

GenaCL600 [577]

So, the force of gravity that the asteroid and the planet have on each other approximately $\boxed{\sf{2.9 \times 10^{17} \: N}}$

<h3>Introduction</h3>

Hi ! Now, I will help to discuss about the gravitational force between two objects. The force of gravity is not affected by the radius of an object, but radius between two object. Moreover, if the object is a planet, the radius of the planet is only to calculate the "gravitational acceleration" on the planet itself,does not determine the gravitational force between the two planets. For the gravitational force between two objects, it can be calculated using the following formula :

$\boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}}$

With the following condition :

F = gravitational force (N)
G = gravity constant ≈ $\sf{6.67 \times 10^{-11}}$ N.m²/kg²
$\sf{m_1}$ = mass of the first object (kg)
$\sf{m_2}$ = mass of the second object (kg)
r = distance between two objects (m)

<h3>Problem Solving</h3>

We know that :

G = gravity constant ≈ $\sf{6.67 \times 10^{-11}}$ N.m²/kg²
$\sf{m_X}$ = mass of the planet X = $\sf{1.55 \times 10^{22}}$ kg.
$\sf{m_Y}$ = mass of the planet Y = $\sf{3.95 \times 10^{28}}$ kg.
r = distance between two objects = $\sf{3.75 \times 10^{11}}$ m.

What was asked :

F = gravitational force = ... N

Step by step :

$\sf{F = G \times \frac{m_X \times m_Y}{r^2}}$

$\sf{F = 6.67 \cdot 10^{-11} \times \frac{1.55 \cdot 10^{22} \cdot 3.95 \times 10^{28}}{(3.75 \times 10^{11})^2}}$

$\sf{F \approx \frac{40.84 \times 10^{-11 + 22 + 28}}{14.0625 \times 10^{22}}}$

$\sf{F \approx 2.9 \times 10^{39 - 22}}$

$\sf{F \approx 2.9 \times 10^{17} \: N}$

<h3>Conclusion</h3>

So, the force of gravity that the asteroid and the planet have on each other approximately

$\boxed{\sf{2.9 \times 10^{17} \: N}}$

Gravity is a thing has depends on ... brainly.com/question/26485200

8 0

1 year ago

A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x (F

Marrrta [24]

Answer:

Part a)

$x = 0.4 m$

Part b)

$v_i = 11.7 m/s$

Part c)

Speed is more than the required speed so it will reach the top

Explanation:

Part a)

As we know that there is no frictional force while block is moving on horizontal plane

so we can use energy conservation on the block

$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$

$\frac{1}{2}0.500(12^2) = \frac{1}{2}(450)x^2$

$x = 0.4 m$

Part b)

If the track has average frictional force of 7 N then work done by friction while block slides up is given as

$W_f = -7( \pi R)$

$W_f = -7(\pi \times 1.00)$

$W_f = -22 J$

work done against gravity is given as

$W_g = - mg(2R)$

$W_g = -(0.500)(9.8)(2\times 1)$

$W_g = -9.8 J$

Now by work energy equation we have

$\frac{1}{2}mv_i^2 + W_f + W_g = \frac{1}{2}mv_f^2$

$\frac{1}{2}0.5(12^2) - 9.8 - 22 = \frac{1}{2}(0.5)v_f^2$

$v_f = 4.1 m/s$

Part c)

now minimum speed required at the top is such that the normal force must be zero

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = 3.13 m/s$

so here we got speed more than the required speed so it will reach the top

5 0

3 years ago

The charge on the sphere is monitored as a beam of monochromatic light shines on the sphere. Initially nothing happens. The wave

Solnce55 [7]

Answer:

explanation of this effect is the photoelectric effect

Explanation:

Let's describe the process, when light of large wavelength falls, this implies a small energy, according to Planck's equation

E = h f = $\frac{h \ c}{ \lambda}$

the energy of the photons is not enough to carry out an electronic transition between two states of the material, when we decrease the wavelength (the energy of the photons increases), the point is reached where the energy of the beam is equal to some energy of a transition, by which the electrons are promoted and since we can see a certain charge, as the atoms are neutral, some electrons must be removed from the material, this is represented in the macroscopic case as the work function of the material, consequently a unbalanced load that is what we can measure.

When we increase the lightning intensity, what we do is that we increase the number of photons and if each photon can remove an electron, by removing the electrons the difference between it and the positive charge (fixed in the nuclei) increases.

We can analyze the interaction of the photon and the electron as a particular collision.

The explanation of this effect was made by Einstein in his explained of the photoelectric effect

8 0

2 years ago

A common flashlight bulb is rated at 0.23 a and 2.9 v (the values of the current and voltage under operating conditions). if the

allochka39001 [22]

The resistance at operating temperature is R = V/I = 2.9 V / 0.23A = 12.61 ohmsT from R – R0 = Roalpha (T – T0), we find that:T = T0 + 1/alpha (R/R0 -1) = 20 degrees Celsius + (1/ 4.3 x 10^-3/K) (12.61 ohms/ 1.1 ohms – 1)T = 2453.40 degrees Celsius

6 0

3 years ago