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Ksju [112]
2 years ago
12

A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 6.0 oz. Aluminum has a density of 2.70 g/cm3. Wh

at is the approximate thickness of the foil in millimeters? Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
SIZIF [17.4K]2 years ago
3 0

Answer:

1.36 x 10^-3 cm

Explanation:

Area = 50 ft^2 = 46451.5 cm^2

mass = 6 oz = 170.097 g

density = 2.70 g/cm^3

Let t be the thickness of foil in cm.

mass = volume x density

mass = area x thickness x density

170.097 = 46451.5 x t x 2.70

t = 1.36 x 10^-3 cm

Thus, the thickness of aluminium foil is 1.36 x 10^-3 cm.

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A jogger runs at a constant rate of 10.0 m every 2.0 seconds. The jogger starts at the origin and runs in the positive direction
Elis [28]

Answer:

(a) 25 m

(b) 75 m

Explanation:

Given that the jogger runs at a constant rate of 10.0 m every 2.0 seconds.

So, the speed of the jogger,

v=\frac{10}{2}=5m/s\;\cdots(i)

Let d be the distance covered by him in time, t s.

As distance=(speed) x (time)

So, d=vt

From equation (i)

\Rightarrow d=5t\;\cdots(ii)

As the jogger starts from origin, so, the distance, d, also represents the position of the jogger at the time t s.

The position-time graph has been shown.

(a) From equation (ii), for t=5.0 s

d=5\times 5=25 m

So, the jogger is at a distance of 25 m from the origin.

(b) Similarly, for t=15.0 s

d=5\times 15=75 m

So, the jogger is at a distance of 75 m from the origin.

8 0
2 years ago
Which would likely demagnetize a magnetized iron nail?
Delicious77 [7]
Dropping it on a hard surface. 
5 0
3 years ago
Read 2 more answers
During lightning strikes from a cloud to the ground, currents as high as 2.50×104 A can occur and last for about 40.0 μs . How m
Nezavi [6.7K]

Answer:

<h3>The charge transferred from the cloud to earth is 1 Coulomb.</h3>

Explanation:

Given :

Current I = 2.5 \times 10^{4} A

Time t = 40 \times 10^{-6} sec

We know that the current is the rate of flow of charge.

From the formula of current,

<h3>  I = \frac{Q}{t}</h3>

Where Q = charge transfer between cloud and earth.

 Q =I t

Q = 2.5 \times 10^{4} \times 40 \times 10^{-6}

Q = 1 C

Hence, the charge transferred from the cloud to earth is 1 Coulomb.

3 0
3 years ago
Why is a specimen smaller than 200 nm not visible with a light microscope?
kifflom [539]
Because the specimen is very small with a light microscope
8 0
2 years ago
In 5 meters, a person running at 0.8 m/s accelerates at 1.6 m/s2. How fast 16 points
frutty [35]

They were going at a velocity 4.07m/s

<u>Explanation:</u>

Distance s =5 m

initial velocity u= 0.8 m/s

Acceleration a =1.6m/s2

We have to calculate the velocity with which they were going afterwards i.e final velocity.

Use the equation of motion

v^2=u^2+2as\\=0.8^2+2\times 1.6\times 5\\=16.64\\v=4.07 m/s

They were going with a velocity 4.07 m/s afterwards.

5 0
3 years ago
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