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Tatiana [17]
3 years ago
11

Can y’all please help

Physics
1 answer:
TiliK225 [7]3 years ago
5 0

Answer:

<em>The cat's speed when it slid off the table was 2.53 m/s</em>

Explanation:

<u>Horizontal Motion</u>

When an object is thrown horizontally at a speed v, from a height h, it describes a curved path ruled by gravity, until it hits the ground.

the range or maximum horizontal distance traveled by the object can be calculated as follows:

\displaystyle d=v\cdot\sqrt{\frac  {2h}{g}}

The situation described in the problem fits into the concept of a horizontal launch when the cat slides off the table and gets to the ground some distance ahead.

It's given the range as d=1.3m and the height h=1.3 m, thus we can find the cat's speed by solving the equation for v:

\displaystyle v=d\cdot\sqrt{\frac  {g}{2h}}

Where g=9.81\ m/s^2

Substituting values:

\displaystyle v=1.3\cdot\sqrt{\frac  {9.81}{2\cdot 1.3}}

\displaystyle v=1.3\cdot\sqrt{\frac  {9.81}{2.6}}

Calculating:

v = 2.53 m/s

The cat's speed when it slid off the table was 2.53 m/s

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A 30-kg child sits at the top of a 3-meter slide. After sliding down, the child is traveling 4 m/s. How much PE does he start wi
Jobisdone [24]
At the top:

         Potential Energy = (mass) x (gravity) x (height)

                                       = (30 kg) x (9.8 m/s²) x (3 meters)

                                       =      882 joules

At the bottom:

           Kinetic Energy  =  (1/2) x (mass) x (speed)²

                                       = (1/2) x (30 kg) x (3 m/s)²

                                       =        (15 kg)  x  (9 m²/s²)

                                       =              135 joules .

He had  882 joules of potential energy at the top,
but only  135 joules of kinetic energy at the bottom.

Friction stole  (882 - 135) = 747 joules of his energy while he slid down.
The seat of his jeans must be pretty warm.
6 0
3 years ago
On a frictionless surface how much force is necessary to accelerate a 0.49 kg object to the left at 4.8 m/s2?
Alenkasestr [34]

Answer:

<h2>2.35 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

force = 0.49 × 4.8 = 2.352

We have the final answer as

<h3>2.35 N</h3>

Hope this helps you

8 0
3 years ago
A bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes.
zaharov [31]

Given that,

Frequency emitted by the bat, f = 47.6 kHz

The speed off sound in air, v = 413 m/s

We need to find the wavelength detected by the bat. The speed of a wave is given by formula as follows :

v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{413}{47.6\times 10^3}\\\\\lambda=0.00867\ m

or

\lambda=8.67\ mm

So, the bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes i.e. 8.67 mm.

3 0
3 years ago
An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r
bixtya [17]

1) Linear charge density of the shell:  -2.6\mu C/m

2)  x-component of the electric field at r = 8.7 cm: 1.16\cdot 10^6 N/C outward

3)  y-component of the electric field at r =8.7 cm: 0

4)  x-component of the electric field at r = 1.15 cm: 1.28\cdot 10^7 N/C outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

1)

The linear charge density of the cylindrical insulating shell can be found  by using

\lambda_2 = \rho A

where

\rho = -567\mu C/m^3 is charge volumetric density

A is the area of the cylindrical shell, which can be written as

A=\pi(b^2-a^2)

where

b=4.7 cm=0.047 m is the outer radius

a=2.7 cm=0.027 m is the inner radius

Therefore, we have :

\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m

 

2)

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

E=E_1+E_2

where:

E_1=\frac{\lambda_1}{2\pi r \epsilon_0}

where

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

E_2=\frac{\lambda_2}{2\pi r \epsilon_0}

where

\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m

And this field points radially inward, because the charge is negative.

Therefore, the net field is

E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C

in the outward direction.

3)

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

E_y=0

4)

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

E=\frac{\lambda_1}{2\pi \epsilon_0 r}

where:

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C

5)

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0
3 years ago
Oil tends to float on water because the density of oil is _____ the density of water.
Rama09 [41]
Oil is less dense than water so it tends to float on the top of the water. Hope this Helps!
7 0
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