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3 years ago

What is the net force acting on a 0.15 hockey puck accelerating at 12 m/s/s?

Physics

1 answer:

svetoff [14.1K]3 years ago

3 0

1.8N

Explanation:

Given parameters:

Mass of the hockey puck = 0.15kg

Acceleration = 12m/s²

Unknown:

Net force = ?

Solution:

According newton's second law of motion "the net force on a body is the product of its mass and acceleration".

Force = mass x acceleration

Input the variables;

Net force = 0.15 x 12 = 1.8N

learn more:

Newton's law of motion brainly.com/question/11411375

#learnwithBrainly

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(12 points) Analysis from the point where the block is released to the point where it reaches the maximum height i) Calculate th

ycow [4]

Answer:

i) a₁ = -g (sin θ + μ cos θ), x = v₀² / 2a₁

ii) W = mg L sin θ , iii) Wₙ = 0

iv) W = - μ m g L cos θ x

Explanation:

With a drawing this exercise would be clearer, I understand that you have a block on a ramp and it is subjected to some force that makes it rise, for example the tension created by a descending block.

The movement is that when the system is released, the tension forces are greater than the friction and the component of the weight and therefore the block rises up the ramp

At some point the tension must become zero, when the hanging block reaches the ground, as the block has a velocity it rises with a negative acceleration to a point and stops where the friction force and the weight component would be in equilibrium along the way. along the plane

i) Let's use Newton's second law

the reference system is with the x axis parallel to the ramp

Axis y

N - W cos θ = 0

X axis

T - W sin θ - fr = ma

the friction force is

fr = μ N

fr = μ mg cos θ

we substitute

T - m g sin sin θ - μ mg cos θ = m a

a = T / m - g (sin θ + μ cos θ)

With this acceleration we can find the height that the block reaches, this implies that at some point the tension becomes zero, possibly when a hanging block reaches the floor.

T = 0

a₁ = -g (sin θ + μ cos θ)

v² = v₀² - 2a1 x

v = 0 at the highest point

x = v₀² / 2a₁

ii) the work of the gravitational force is

W = F .d

W = mg sin θ L

iii) the work of the normal force

the force has 90º with respect to the displacement so cos 90 = 0

Wₙ = 0

iv) friction force work

friction force always opposes displacement

W = - fr d

W = - μ m g cos θ L

4 0

3 years ago

Kjebhg da sad bmm m mmm b bc f gaf see Krio ok pop it ya ta is v BSA u lie lug

GenaCL600 [577]

Answer: Kjebhk das dajsd ajdn wadm awkdlwandeA fefme ef

4 0

3 years ago

Drops of rain fall perpendicular to the roof of a parked car during a rainstorm. The drops strike the roof with a speed of 15 m/

IrinaK [193]

Answer: (a) 1.065 N (b) 2.13 N

Explanation:

<h2>(a) average force exerted by the rain on the roof</h2><h2 />

According Newton's 2nd Law of Motion the force $F$ is defined as <u>the variation of linear momentum</u> $p$ <u>in time:</u>

$F=\frac{dp}{dt}$ (1)

Where the linear momentum is:

$p=mV$ (2) Being $m$ the mass and $V$ the velocity.

In the case of the rain drops, which initial velocity is $V_{i}=15m/s$ and final velocity is $V_{f}=0$ (we are told the drops come to rest after striking the roof). The momentum of the drops $p_{drops}$ is:

$p_{drops}=mV_{i}+mV_{f}$ (3)

If $V_{f}=0$ , then:

$p_{drops}=mV_{i}$ (4)

Now the force $F_{drops}$ exerted by the drops is:

$F_{drops}=\frac{dp_{drops}}{dt}=\frac{d}{dt}mV_{i}$ (5)

$F_{drops}=\frac{dm}{dt}V_{i}+m\frac{dV_{i}}{dt}$ (6)

At this point we know the mass of rain per second (mass rate) $\frac{dm}{dt}=0.071 kg/s$ and we also know the initial velocity does not change with time, because that is the velocity at that exact moment (instantaneous velocity). Therefore is a constant, and the derivation of a constant is zero.

This means (6) must be rewritten as:

$F_{drops}=\frac{dm}{dt}V_{i}$ (7)

$F_{drops}=(0.071 kg/s)(15m/s)$ (8)

$F_{drops}=1.065kg.m/s^{2}=1.065N$ (9) This is the force exerted by the rain drops on the roof of the car.

<h2>(b) average force exerted by hailstones on the roof </h2><h2 />

Now let's assume that instead of rain drops, hailstones fall on the roof of the car, and let's also assume these hailstones bounce back up off after striking the roof (this means they do not come to rest as the rain drops).

In addition, we know the hailstones fall with the same velocity as the rain drops and have the same mass rate.

So, in this case the linear momentum $p_{hailstones}$ is:

$p_{hailstones}=mV_{i}+mV_{f}$ (9) Being $V_{i}=V_{f}$

$p_{hailstones}=mV+mV=2mV$ (10)

Deriving with respect to time to find the force $F_{hailstones}$ exerted by the hailstones:

$F_{hailstones}=\frac{d}{dt}p_{hailstones}=\frac{d}{dt}(2mV)$ (10)

$F_{hailstones}=2\frac{d}{dt}(mV)=2(\frac{dm}{dt}V+m\frac{dV}{dt})$ (11)

Assuming $\frac{dV}{dt}=0$ :

$F_{hailstones}=2(\frac{dm}{dt}V)$ (12)

$F_{hailstones}=2(0.071 kg/s)(15m/s)$ (13)

Finally:

$F_{hailstones}=2.13kg.m/s^{2}=2.13N$ (14) This is the force exerted by the hailstones

Comparing (9) and (14) we can conclude the force exerted by the hailstones is two times greater than the force exerted by the raindrops.

5 0

3 years ago

A common laboratory reaction is the neutralization of an acid with a base. When 50.0 mL of 0.500 M HCl at 25.0°C is added to 50.

GalinKa [24]

Answer:

Explanation:

In this calorimetry problem, the heat released by the reaction is equal to the heat absorbed by the solution (assumed to have the same specific heat capacity as water, 4.19 Jg⁻¹°C⁻¹).

The formula Q = mcΔt will be used to calculate the heat energy, where m is the mass, c is the specific heat capacity, and Δt is the change in temperature from final to initial.

The volume of solution is (50.0 + 50.0)mL = 100.0mL = 100.0g, since water has a density of 1.00g/mL.

The heat absorbed by the solution is then calculated.

Q = mcΔt = (100.0 g)(4.19 Jg⁻¹°C⁻¹)(28.2°C - 25.0°C) = 1340 J

The closest answer is B) 1300 J. This answer is obtained by including only two significant figures in the answer.

7 0

3 years ago

A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache

Paladinen [302]

Answer:

$F_a=1470\ N$

Explanation:

<u>Friction Force</u>

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

$\displaystyle F_a-F_{r1}-F_{r2}=m.a=0$

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

$\displaystyle F_a=F_{r1}+F_{r2}.....[1]$

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

$\displaystyle F_{r2}-T=0$