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3 years ago

Which of these letters is the symbol for current in equations A: c B: i C: r D: t

1 answer:

4 0

Answer: [B]: The letter, " I " ; for current; in units of "Amps" .
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Newton's law of cooling is , where is the temperature of an object, is in hours, is a constant ambient temperature, and is a pos

kirill [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The time taken is $t = 6.89 \ hrs$

Explanation:

From the question we are told that

The value of $k = 0.15hr^{-1}$

The the interior temperature is $u(t = 0) = 70 ^oF$

The external temperature is $T = 11 ^oF$

The required interior temperature is $u_{t} = 32 ^oF$

The newton cooling law is

$\frac{du}{dt} = -k (u -T)$

=> $\frac{du}{u-T} = - kdt$

Now integrate both sides we have

$\int\limits \frac{du}{u-T} =\int\limits - kdt$

$ln (u - T) = -kt + c$

Since c is a constant lnC = c will also give a constant so

$ln (u - T) = -kt + ln C$

=> $\frac{(u -T)}{C} = e^{-kt}$

=> $u = T + Ce^{-kt}$

substituting value

$70 = 11 +Ce^{-0* 0.15}$

$C = 59$

Hence

$u_t = 11 + 59 e^{-0.15 *t}$

$32= 11 + 59 e^{-0.15 *t}$

=> $t = -\frac{ln(\frac{21}{59} )}{0.15}$

$t = 6.89 \ hrs$

3 0

3 years ago

A train travels 55 kilometers in 2 hours, and then 52 kilometers in 5 hours. What is its average speed?

bulgar [2K]

The formula for working out speed is distance ÷ time.

55 km ÷ 2 hours = 27.5 km/h (average speed for first part of journey)

52km ÷ 5 hours = 10.4 km/h (average speed for second part of journey)

(27.5 + 10.4) ÷ 2 = 18.95 km/h (average speed throughout the journey)

4 0

3 years ago

Read 2 more answers

If your prediction turns out to be false your experiment has been a failure

nikdorinn [45]

This is false. Your hypothesis, or prediction, is just that: a prediction. Saying its a failure will result in bias.

3 0

3 years ago

Read 2 more answers

If a 20 g cannonball is shot from a 5 kg cannon with a velocity of 100

balu736 [363]

Strange as it may seem, the statement in the question appears to be TRUE.

-- Before the shot, neither the cannon nor the ball is moving, so their combined momentum is zero.

-- Since momentum is conserved, we know immediately that their combined momentum AFTER the shot also has to be zero.

-- (20g is rather puny for a "cannonball" ... about the same weight as four nickels. But we'll take your word for it and just do the Math and the Physics.)

-- Momentum = (mass) x (velocity)

After the shot, the momentum of the cannonball is

(0.02 kg) x (100 m/s ==> that way)

Momentum of the ball = 2 kg-m/s ==> that way.

-- In order for both of them to add up to zero, the momentum of the cannon must be (2 kg-m/s this way <==) .

Momentum of cannon = (5 kg) x (V m/s this way <==)

2 kg-m/s this way <== = (5 kg) x (V m/s this way <==)

Divide each side by (5 kg):

V m/s = (2/5) m/s this way <==

Speed of recoil of the cannon = -- 0.4 m/s

3 0

3 years ago

The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a pus

hoa [83]

Answer:

Incomplete question: "Each block has a mass of 0.2 kg"

The speed of the two-block system's center of mass just before the blocks collide is 2.9489 m/s

Explanation:

Given data:

θ = angle of the surface = 37°

m = mass of each block = 0.2 kg

v = speed = 0.35 m/s

t = time to collision = 0.5 s

Question: What is the speed of the two-block system's center of mass just before the blocks collide, vf = ?

Change in momentum:

$delta(P)=F*delta(t)$

$P_{f} -P_{i}=F*delta(t)$

$2m(v_{f} -v_{i})=F*delta(t)$

$v_{i} =0.35-0.35=0$

It is neccesary calculate the force:

$F=(m+m)*g*sin\theta$

Here, g = gravity = 9.8 m/s²

$F=(0.2+0.2)*9.8*sin37=2.3591N$

$v_{f} =\frac{F*delta(t)}{2m} =\frac{2.3591*0.5}{2*0.2} =2.9489m/s$

6 0

3 years ago