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3 years ago

How many oranges did Kevin buy if he has three bags full and one bag can hold 76 oranges?

Physics

2 answers:

seraphim [82]3 years ago

8 0

Answer:

228

Explanation:

:)I hope you enjoy this:)

76*3=228

Mrac [35]3 years ago

6 0

I believe the answer would be 228

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Suppose astronomers built a 20-meter telescope. How much greater would its light-collecting area be than that of the 10-meter Ke

nirvana33 [79]

Answer:

4 times greater

Explanation:

Step 1: Calculate light-collecting area of a 20-meter telescope (A₁) by using area of a circle.

Area of circle = π*r² = $\frac{\pi d^{2}}{4}$

Where d is the diameter of the circle = 20-m

$A_{1} = \frac{\pi d^{2}}{4}$

$A_{1} = \frac{\pi (20^{2})}{4}$

A₁ = 314.2 m²

Step 2: Calculate light-collecting area of a 10-meter Keck telescope (A₂)

$A_{2} = \frac{\pi d^{2}}{4}$

Where d is the diameter of the circle = 10-m

$A_{2} = \frac{\pi (10^{2})}{4}$

A₂ = 78.55 m²

Step 3: divide A₁ by A₂

$= \frac{314.2 m^2}{78.55 m^2}$

= 4

Therefor, the 20-meter telescope light-collecting area would be 4 times greater than that of the 10-meter Keck telescope.

5 0

3 years ago

An object initially at rest experiences an acceleration of 9.8m/s2. How much time will it take it to achieve a velocity of 49m/s

Vlad1618 [11]

Answer:5seconds

Explanation:

initial velocity(u)=0

Final velocity(v)=49m/s

acceleration(a)=9.8m/s2

Time(t)=?

V=u+axt

49=0+9.8xt

49=9.8t

t=49/9.8

t=5seconds

6 0

3 years ago

What would happen if there were no enzymes in the human body?

MakcuM [25]

Answer:

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3 years ago

Read 2 more answers

A pipe open only at one end has a fundamental frequency of 266 Hz. A second pipe, initially identical to the first pipe, is shor

Alika [10]

Answer:

1.16cm were cut off the end of the second pipe

Explanation:

The fundamental frequency in the first pipe is,

Since the speed of sound is not given in the question, we would assume it to be 340m/s

f1 = v/4L, where v is the speed of sound and L is the length of the pipe

266 = 340/4L

L = 0.31954 m = 0.32 m

It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider L’ be the length that was cut from the first pipe.

So, the length of the second pipe is L – L’

Then, the fundamental frequency in the second pipe is

f2 = v/4(L - L’)

The beat frequency due to the fundamental frequencies of the first and second pipe is

f2 – f1 = 10hz

[v/4(L - L’)] – 266 = 10

[v/4(L – L’)] = 10 + 266

[v/4(L – L’)] = 276

(L - L’) = v/(4 x 276)

(L – L’) = 340/(4 x 276)

(L – L’) = 0.30797

L’ = 0.31954 – 0.30797

L’ = 0.01157 m = 1.157 cm ≅ 1.16cm

Hence, 1.16 cm were cut from the end of the second pipe

6 0

3 years ago

A boat with a mass of 1000 kg drifts with the current down a straight section of river parallel to the +x+x axis with a speed of

IrinaK [193]

Answer:

Explanation:

Our values are defined by,

$m=1000kg\\v = 2m/s\\a = 2.7 \angle 45\°$

We can express also in a vectorial way, then

$v=2\hat{i} \rightarrow$ no values in $\hat{j}$

Acceleration as follow,

$a= 2.7cos45 \hat{i} + 2.7 sin45 \hat{j}$

We know that velocity is given by,

$V(t) = v_0+at \rightarrow v_0 = 2$

We need to calculate for t=3, then

$v(3) = 2\hat{i} +(2.7cos45 \hat{i} + 2.7 sin45 \hat{j})*3$

$v(3) = (2\hat{i}+3*2.7cos45 \hat{i})+(3*2.7 sin45 \hat{j})$

$v(3) = (2+3*2.7cos45)\hat{i}+(5.7 sin45 )\hat{j}$

$v(3) = 7.7\hat{i}+5.7\hat{j}$

Our mass is 1000Kg, so the momentum is

$P = mv$

$P= 1000(7.7\hat{i}+5.7\hat{j})$

$P=7700\hat{i}+5700\hat{j}$

7 0

4 years ago