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4 years ago

This is an example of using a hammer as a(n)

Physics

2 answers:

barxatty [35]4 years ago

3 0

, to hit nails into a piece of wood or a wall, or to break things into pieces.

Burka [1]4 years ago

3 0

Answer:

It's A

Explanation:

the answer on e2020 is A, a lever

You might be interested in

small plastic container, called the coolant reservoir, catches the radiator fluid that overflowswhen the automobile engine becom

Ilia_Sergeevich [38]

Answer:

0.53 quart

Explanation:

The volume expansion of the coolant is gotten from ΔV = VγΔθ where ΔV = change in volume of the coolant, V = initial volume of coolant = 15 quart, γ = coefficient of volume expansion of coolant = 410 × 10⁻⁶ /°C and Δθ = temperature change = θ₂ - θ₁ where θ₁ = initial temperature of coolant = 6 °C and θ₂ = final temperature of coolant = 92 °C. So, Δθ = θ₂ - θ₁ = 92 °C - 6 °C = 86 °C

Since, ΔV = VγΔθ

substituting the values of the variables into the equation, we have

ΔV = VγΔθ

ΔV = 15 × 410 × 10⁻⁶ /°C × 86 °C

ΔV = 528900 × 10⁻⁶ quart

ΔV = 0.528900 quart

ΔV ≅ 0.53 quart

Since the change in volume of the coolant equals the spill over volume, thus the overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92 °C is 0.53 quart.

4 0

3 years ago

If Rebecca is traveling at 7 m/s east for 1 hour how much is Rebecca displaced

Mekhanik [1.2K]

Answer:

25.2 km east

Explanation:

The relationship between velocity and displacement is given by:

$v=\frac{d}{t}$

where

v is the velocity

d is the displacement

t is the time

In this problem, we have:

$v=7 m/s$ east is Rebecca's velocity

$t=1 h=60 min=3600 s$ is the time taken

Re-arranging the equation, we can find Rebecca's displacement:

$d=vt=(7 m/s)(3600 s)=25,200 m=25.2 km$

and the direction is same as velocity (east)

3 0

3 years ago

A ball hits a wall. What is true about the magnitude of the force experienced by the ball compared with the force experienced by

Ivahew [28]

<span>This is best understood with Newtons Third Law of Motion: for every action there is an equal and opposite reaction. That should allow you to see the answer.</span>

5 0

3 years ago

A 700-kg car, driving at 29 m/s, hits a brick wall and rebounds with a speed of 4.5 m/s. what is the car's change in momentum du

Viefleur [7K]

The change in the momentum of the car due to collision between car and wall is $\fbox{\begin\ -23450\text{ kg.m}/\text{s}\end{minispace}}$ or $\fbox{\begin\\-2.3450 \times {10^4}\,\text{kg.m}/\text{s}\end{minispace}}$ .

Further Explanation:

Let us consider the car is moving towards the right direction and it collides with the wall. After the collision, the car will bounce back or car will rebound opposite to the direction of its motion i.e., towards left. Therefore, the final velocity of the car is opposite to the direction of the initial velocity of the car.

Given:

The mass of the car is $700\,{\text{kg}}$ .

The velocity of the car before collision is $29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

The velocity of the car after collision is $4.5\text{ m}/\text{s}$ .

Concept:

The momentum of an object is defined as the product of mass of object and the velocity with which the object is moving.

The initial momentum of the car is:

$\fbox{\begin\\{p_i}= m{v_i}\end{minispace}}$ …… (I)

Here, ${p_i}$ is the initial momentum of the car, $m$ is the mass of the car and ${v_i}$ is the initial velocity of the car.

Substitute $700\,{\text{kg}}$ for $m$ and $29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${v_i}$ in equation (I).

$\begin{gathered}{p_i} = \left( {700\,{\text{kg}}} \right) \cdot \left( {29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}} \right) \\ = 20300\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\ \end{gathered}$

The final momentum of the car is defined as the product of mass of car and the velocity of the car after collision or final velocity of the car.

The final momentum of the car is:

$\fbox{\begin\\{p_f} = m{v_f}\end{minispace}}$ …… (II)

Here, ${p_f}$ is the final momentum and ${v_f}$ is the final velocity.

Substitute $700\,{\text{kg}}$ for $m$ and - $4.5\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${v_f}$ in equation (II).

$\begin{aligned}{p_f}&=\left( {700\,{\text{kg}}} \right)\cdot\left( {- 4.5\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}}{\text{s}}}} \right.\kern-\nulldelimiterspace}{\text{s}}}}\right)\\&=-3150\,{{{\text{kg}} \cdot {\text{m}}}\mathord{\left/{\vphantom{{{\text{kg}}\cdot {\text{m}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

The change in the momentum of the car after collision is the difference between the momentum of car before collision and the momentum of car after collision.

The change in momentum of the car is:

$\fbox{\begin\Delta p = {p_f} - {p_i}\end{minispace}}$ …… (III)

Here, $\Delta p$ is the change in the momentum of the car.

Substitute $- 3150\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${p_f}$ and $20300\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${v_i}$ in equation (III).

$\begin{aligned}\Delta{p}&=-3150\text{ kg}.\text{m}/\text{s}\ -20300\text{ kg}.\text{m}/\text{s}\\&=-23450\text{ kg}.\text{m}/\text{s}\end{aligned}$

Thus, the change in the momentum of the car due to collision between car and wall is $\fbox{\begin\ -23450\text{ kg.m}/\text{s}\end{minispace}}$ or $\fbox{\begin\\-2.3450 \times {10^4}\,\text{kg.m}/{s}\end{minispace}}$ .

Learn more:

1. The motion of a body under friction brainly.com/question/4033012/

2. A ball falling under the acceleration due to gravity brainly.com/question/10934170/

3.Conservation of energy brainly.com/question/3943029/

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Change in momentum, collision, initial velocity, final velocity, initial momentum, final momentum,-23450 kgm/s^2, -23450 kgm/s2, -2.3450*10^6 kgm/s^2, -2.3450*10^6 kgm/s2.

7 0

3 years ago

I need answers and solvings to these questions

den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

5 0

3 years ago