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2 years ago

If Juan upgrades from a sports car to a large truck, what will happen to the

Physics

1 answer:

vovikov84 [41]2 years ago

4 0

Answer:

The force will have to increase

Explanation:

Since Juan has upgraded from a sports car to a large truck, based on Newton's second law of motion, the force needed to keep the truck going at the same speed will have to increase.

According to Newton's second law "the force on an object is equal to the product of its mass and acceleration".

Force = mass x acceleration

A truck has a larger mass compared to a sports car.

By virtue of this, to make sure both automobiles attain the same speed, the force powering them to accelerate must be the same.

Therefore, the force from the engine must increase.

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Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a

ad-work [718]

Answer:

Maximum speed of the car is 17.37 m/s.

Explanation:

Given that,

Radius of the circular track, r = 79 m

The coefficient of friction, $\mu=0.39$

To find,

The maximum speed of car.

Solution,

Let v is the maximum speed of the car at which it can safely travel. It can be calculated by balancing the centripetal force and the gravitational force acting on it as :

$v=\sqrt{\mu rg}$

$v=\sqrt{0.39\times 79\times 9.8}$

v = 17.37 m/s

So, the maximum speed of the car is 17.37 m/s.

6 0

3 years ago

A spring does 5.0 J of work on a 0.10-kg ball bearing in a pinball machine. The ball's

lisov135 [29]

Answer:

10m/s

Explanation:

5 0

3 years ago

Differences between <br>hor<br>rse<br>and horse

kenny6666 [7]

The proper difference between hor Rse and horse is both shows the same thing that is the horse

5 0

3 years ago

Read 2 more answers

A spring of spring constant k is attached to a support at the bottom of a ramp that makes an angle θ with the horizontal. A bloc

Nikitich [7]

Answer:

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

Explanation:

From the law of conservation of energy

Energy lost by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr

$0.5kd^{2}=0.5mv^{2}+mgLsin\theta+\mu_{k}mgcos\theta x$

$x(mgsin\theta+\mu_{k}mgcos\theta)=0.5kd^{2}$

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

The required distance from A to B is $x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

5 0

3 years ago

An electron moves through a uniform electric field vector E = (2.80î + 5.20ĵ) V/m and a uniform magnetic field vector B = 0.400k

alina1380 [7]

Answer:

1.758820×10^11(-2.5i-0.8j) m/s^2

Explanation:

From the question, the parameters given are; E=(2.80i+ 5.20j) v/m, a uniform magnetic field,B= 0.400K T, acceleration, a= ??? and velocity vector, v= 11.0i metre per seconds (m/s)...

We can solve this problem using the formula below;

Ma= q[E+V × B] ---------------(1).

Note: q is negative, m= mass of electron.

Making acceleration,a the subject of the formula and substituting the parameters into equation (1);

a= -e/m × (2.5i + 5.2j +11.0i × 0.400K)

a= -e/m × (2.5i+5.2j-4.4j)

a= e/m × (-2.5i - 0.8j)

e/m= 1.758820×10^11 c/kg

Therefore, slotting in the value of charge to mass(e/m) ratio;

a= 1.7588×10^11×(-2.5i-0.8j) m/s^2

7 0

3 years ago