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3 years ago

Use newton's law to explain the vertical acceleration of a projectile

Physics

1 answer:

NISA [10]3 years ago

5 0

Answer:

Explained below

Explanation:

Newton's first law of motion: This law states that an object will remain at rest or continue in constant motion except it's acted upon by an external force. In projectile motion, the horizontal component of velocity will remain unchanged because we ignore air resistance since no force is acting in that horizontal direction.

Newton's second law of motion: This law states that force is the product of mass and acceleration. In projectile the force acts downwards, thus f = mg.

But g = a since internal forces will cancel out.

Thus, F = ma

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A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on

MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

$Ec=\frac{1}{2} *m*v^{2}$

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

$W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}$

$W=\frac{1}{2} *m*(v2^{2}-v1^{2})$

In this case:

W=?
m= 2,145 kg
v2= 12 $\frac{m}{s}$

v1= 25 $\frac{m}{s}$

Replacing:

$W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})$

W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

3 0

3 years ago

A very long solid insulating cylinder has radius R = 0.1 m and uniform charge density rho0= 10-3 C/m3. Find the electric field a

Galina-37 [17]

Answer:

$E = (0.56 \times 10^8 ) r \ \ N/c$

Explanation:

Given that:

$\rho_o = (10^{-3} ) \ c/m^3$

R = (0.1) m

To find the electric field for r < R by using Gauss Law

${\oint}E^{\to}* da^{\to} = \dfrac{Q_{enclosed}}{\varepsilon_o} --- (1)$

For r < R

$Q_{enclosed}=(\rho) ( \pi r^2 ) l$

$E*(2 \pi rl)= \dfrac{\rho ( \pi r ^2 l)}{\varepsilon_o}$

$E= \dfrac{\rho ( r)}{2 \varepsilon_o}$

where;

$\varepsilon_o = 8.85 \times 10^{-12}$

$E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}$

$E = (0.56 \times 10^8 ) r \ \ N/c$

4 0

3 years ago

If a car increases its velocity from zero to 60 m/s in 10 seconds, its acceleration is

andrezito [222]

We know that a=vf_vi/t equals equation "a" . Where a is the acceleration of the body , vf is the final velocity , vi is the initial velocity and t is equal to time . Since vi equals o m/s , vf equals to 60 m/s and t equals 10 s. Put in equation "a". a=60-0/10 =6m/s2

5 0

3 years ago

Helpp me pleassee....

lilavasa [31]

Answer:

The fundamental wavelength of the vibrating string is 1.7 m.

Explanation:

We have,

Velocity of wave on a guitar string is 344 m/s

Length of the guitar string is 85 cm or 0.85 m

It is required to find the fundamental wavelength of the vibrating string. The fundamental frequency on the string is given by :

$f=\dfrac{v}{2l}\\\\f=\dfrac{344}{2\times 0.85}\\\\f=202.35\ Hz$

Now fundamental wavelength is :

$\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{344}{202.35}\\\\\lambda=1.7\ m$

So, the fundamental wavelength of the vibrating string is 1.7 m.

4 0

3 years ago

What is the KE if a 10kg mass traveling at 5 m/s

11111nata11111 [884]

Answer: KE = 25 J

Explanation: You must use the formula

KE = 1/2 m v²

to solve this problem.

KE = 1/2 (10 Kg) (5 m/s)

KE = 1/2 (50 kgm/s)

KE = 25 J

7 0

3 years ago