All categories

3 years ago

PLEASE HELP WILL GIVE 10 POINTS!!!!

Physics

2 answers:

Leokris [45]3 years ago

4 0

Answer:

The answer is A

Explanation:

katrin2010 [14]3 years ago

3 0

The correct answer would be A

You might be interested in

Sarah's group designed this oven and eventually melted the

torisob [31]

Answer:

Explanation:

because i don't he asking me

5 0

3 years ago

What is the likely identity of a metal if a sample has a mass of 63.5 g when measured in air and an apparent mass of 60.2 g when

Gre4nikov [31]

Answer:

Gold

Explanation:

Given:

Mass of sample = 63.5 g

Mass of water = 60.2 g

Find:

Object

Computation:

Mass of water displaced = 63.5 g - 60.2 g

Mass of water displaced = 3.3 g

So, volume in water = 3.3 cm³

Density = Mass / Volume

Density = 63.5 g / 3.3

Density = 19.24

So,

Object ,must be gold.

7 0

2 years ago

A rocket rises vertically, from rest, with an acceleration of 3.2 m/s2 until it runs out of fuel at an altitude of 1300 m. After

svetlana [45]

Explanation:

Below is an attachment containing the solution.

6 0

3 years ago

(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

5 0

3 years ago

A rifle of mass M is initially at rest, but is free to recoil. It fires a bullet of mass m with a velocity +v relative to the gr

Natali5045456 [20]

Answer:

$V=-\dfrac{mv}{M}$

Explanation:

Given that

Mass of rifle = M

Initial velocity ,u= 0

Mass of bullet = m

velocity of bullet = v

Lets take final speed of the rifle is V

There is no any external force ,that is why linear momentum of the system will be conserve.

Initial linear momentum = Final linear momentum

M x 0 + m x 0 = M x V + m v

0 = M x V + m v

$V=-\dfrac{mv}{M}$

Negative sign indicates that ,the recoil velocity will be opposite to the direction of bullet velocity.

6 0

3 years ago