Which statement is correct about the equation for work?

What is the Force in Newtons exerted by cart with a mass of 0.75 kg and an Acceleration of 6m/s2?

denis-greek [22]

Answer:

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 we have

force = 0.75 × 6 = 4.5

We have the final answer as

Hope this helps you

6 0

2 years ago

The average kinetic energy of the molecules of an ideal gas is directly proportional to the

Cerrena [4.2K]

The average kinetic energy<span> of a </span>gas<span> particle is </span>directly proportional<span> to the </span>temperature<span>.</span>

3 0

3 years ago

A proud new Jaguar owner drives her car at a speed of 25 m/s into a corner. The coefficients of friction between the road and th

ehidna [41]

Answer:

ac = 3.92 m/s²

Explanation:

In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,

Frictional Force = Centripetal Force

where,

Frictional Force = μ(Normal Force) = μ(weight) = μmg

Centripetal Force = (m)(ac)

Therefore,

μmg = (m)(ac)

ac = μg

where,

ac = magnitude of centripetal acceleration of car = ?

μ = coefficient of friction of tires (kinetic) = 0.4

g = 9.8 m/s²

Therefore,

ac = (0.4)(9.8 m/s²)

5 0

3 years ago

Death Star has a diameter of 160,000m and a mass of 5.1e17kg. Millennium Falcon has a mass of 1.36e6kg (data from Wookieepedia)

lions [1.4K]

Answer:

7229 N

Explanation:

The gravitational force between the Death Star and the Millenium Falcon is given by:

$F=G\frac{mM}{R^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$M=5.1\cdot 10^{17} kg$ is the mass of the Death Star

$m=1.36\cdot 10^6 kg$ is the mass of the Millennium Falcon

$R=\frac{160,000 m}{2}=80,000 m$ is the radius of the Death Star

Substituting numbers into the equation, we find the force

$F=(6.67\cdot 10^{-11})\frac{(1.36\cdot 10^6 kg)(5.1\cdot 10^{17} kg)}{(80,000 m)^2}=7229 N$

5 0

3 years ago

A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the

BartSMP [9]

Answer:

$r=1.14m$

Explanation:

$\theta$ is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

$F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB$

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

$F_m=F_c\\qvB=F_c(1)$

$F_c$ is the centripetal force and is defined as:

$F_c=m\frac{v^2}{r}$

Here $v$ is the proton's speed and $r$ is the radius of the circular motion. Replacing this in (1) and solving for r:

$qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}$

Recall that 1 J is equal to $6.242*10^{12}MeV$ , so:

$4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J$

We can calculate $v$ from the kinetic energy of the proton:

$K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}$

Finally, we calculate the radius of the proton path:

$r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m$

8 0

3 years ago