If a marble is released from a height of 10 m how long would it take for it to hit the ground?

Newton’s empirical law of cooling/warming of an object is given by ( ), T Tm k dt dT = − where k is a constant of proportionalit

pychu [463]

Answer:

The time for the cake to cool off to room temperature is

approximately 30 minutes.

Let $T_{0}$ = $70^{0}$ F be the temperature and T that of the body

Explanation:

Our Tm = 70, the initial-value problem is

$\frac{DT}{dt}$ = k(T − 70), T(0) = 300

Solving the equation, we get

$\frac{DT}{t-70}$ = kdt

In [T-70]= kt + $C_{1}$

T = 70 + $C_{2}$ $e^{kt}$

Finding he value for $C_{2}$ using the initial value of T (0)= 300, therefore we get:

300=70+ $C_{2}$

$C_{2}$ = 230 therefore

T= 70+ 230 $e^{kt}$

Finding the value for k using T (3) = 200, therefore we get

T (3) = 200

$e^{3k}$ = $\frac{13}{23}$

K = $\frac{1}{3}$ in $\frac{13}{23}$

= -0.19018

Therefore

T(t) = 70+230 $e^{-0.19018}$

4 0

3 years ago

Consider a roller coaster begins 15m above the ground. If the cart has a mass of 75kg, what is the velocity of the cart halfway

SashulF [63]

Answer:

v = 12.12 m/s

Explanation:

Given that,

The mass of the cart, m = 75 kg

The roller coaster begins 15 m above the ground.

We need to find the velocity of the cart halfway to the ground. Let the velocity be v. Using the conservation of energy at this position, h = 15/2 = 7.5 m

$mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 7.5} \\\\v=12.12\ m/s$

So, the velocity of the cart is 12.12 m/s.

7 0

3 years ago

Calculate the acceleration of a 1000 kg car if the motor provides a small thrust of 1000 N and the static and dynamic friction c

grin007 [14]

Explanation :

It is given that,

Mass of the car, m = 1000 kg

Force applied by the motor, $F_A=1000\ N$

The static and dynamic friction coefficient is, $\mu=0.5$

Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,

$F_A-\mu mg=ma$

$\dfrac{F_A-\mu mg}{m}=a$

$a=\dfrac{1000-0.5(1000)(9.81)}{1000}$

$a=-3.905\ m/s^2$

So, the acceleration of the car is $-3.905\ m/s^2$ . Hence, this is the required solution.

6 0

3 years ago

Which of the following could you do to increase the strength of an electromagnet?

Ann [662]

Wrap around a metal with wire instead of using wire alone.

8 0

3 years ago

A freight car moves along a friction less level railroad track at constant speed. The car is open on top. A large load of coal i

maxonik [38]

Answer:

The velocity of the freight car decreases.

Explanation:

This question is answered by the conservation of momentum principle.

When the freight car is moving at a certain speed, it has a constant momentum.

We will call this M1.

The equation for M1 will be:

M1 = Mass * Speed

Now when the coal is dumped into the freight car, the Mass increases.

Since conservation of momentum states that the momentum will remain the same. We have:

M1 = (Mass of freight + Mass of coal) * Speed

Since M1 is constant, if the mass increases, the speed had to decrease to keep the equation true.

8 0

3 years ago