A translucent material...

At the moment a hot cake is put in a cooler, the difference between the cakes and the cooler's

hram777 [196]

Answer: D(t)= 50(4/5)^t

Explanation: If 1/5 of the temperature difference is lost each minute, that means 4/5 of the difference remains each minute. So each minute, the temperature difference is multiplied by a factor of 4/5 (or 0.8).

If we start with the initial temperature difference, 50° Celsius, and keep multiplying by 4/5, this function gives us the temperature difference t minutes after the cake was put in the cooler.

5 0

2 years ago

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a concrete slab 20 m long and weighing 400,000 N is supported by one pillar. If a 19,600 N car is parked 8 meters from one end.

LUCKY_DIMON [66]

let the distance of pillar is "r" from one end of the slab

So here net torque must be balance with respect to pillar to be in balanced state

So here we will have

$Mg(r - L/2) = mg(L/2 - 8)$

here we know that

mg = 19600 N

Mg = 400,000 N

L = 20 m

from above equation we have

$400,000(r - 10) = 19,600 (10 - 8)$

$r - 10 = 0.098$

$r = 10.098 m$

so pillar is at distance 10.098 m from one end of the slab

7 0

3 years ago

True or False: The arrows on a motion map should point in the directions of motion.

slavikrds [6]

Answer:

true,true,false

Explanation:

its false because if it is equal it would show an arrow pointing left and a 20 and the same for the right

3 0

2 years ago

A medical cyclotron used in the production of medical isotopes accelerates protons to 6.5 MeV. The magnetic field in the cyclotr

mart [117]

Answer:

diameter of largest orbit is 0.60 m

Explanation:

given data

isotopes accelerates KE = 6.5 MeV

magnetic field B = 1.2 T

to find out

diameter

solution

first we find velocity from kinetic energy equation

KE = 1/2 × m×v² ........1

6.5 × 1.6 × $10^{-19}$ = 1/2 × 1.672 × $10^{-27}$ ×v²

v = 3.5 × $10^{7}$ m/s

so

radius will be

radius = $\frac{m*v}{B*q}$ ........2

radius = $\frac{1.672*10^{-27}*3.5*10^{7}}{1.2*1.6*10^{-19}}$

radius = 0.30

so diameter = 2 × 0.30

so diameter of largest orbit is 0.60 m

8 0

3 years ago

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Si se deja caer un carrito de una pista de coches sin friccion y su altura inicial es de 1.4 metros, cual es la velocidad maxima

Svet_ta [14]

Answer:

$5.241\ \text{m/s}$

Explanation:

m = Masa del coche

g = Aceleración debida a la gravedad = $9.81\ \text{m/s}^2$

h = Altura = $1.4\ \text{m}$

v = Velocidad del automóvil en la parte inferior de la pista

Aquí asumimos que el automóvil desciende verticalmente. La energía potencial del automóvil se completará convertida en energía cinética en la parte inferior de la pista ya que no hay pérdida de energía.

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.4}\\\Rightarrow v=5.241\ \text{m/s}$

La velocidad máxima que puede alcanzar el coche es $5.241\ \text{m/s}$ .

8 0

3 years ago