Write the word AMBULANCE as it would appear when reflected in a plane mirror

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is th

katrin [286]

Complete Question

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long.What Average force is excreted on the Nail

Answer:

$F=2*10^{4}N$

Explanation:

From the question we are told that:

Mass $m=0.500kg$

Initial Velocity $V=15.0m/s$

Distance $x=2.80mm=>0.00280m$

Diameter $d=2.50mm=>0.00250m$

Length $l=6.00cm=>0.6m$

Generally the equation for Force is mathematically given by

$F=\frac{mv^2}{2d}$

$F=\frac{0.500*15^2}{2.80*10^{-3}}$

$F=2*10^{4}N$

6 0

2 years ago

As Aubrey watches this merry-go-round for a total of 2 minutes, she notices the black horse pass by 15 times. What is the period

marishachu [46]

2 min = 120 sec

120/15 = 8

The black horse represents 8 seconds.

8 0

3 years ago

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You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi

Mama L [17]

Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

6 0

3 years ago

At the local swimming hole, a favorite trick is to run horizontally off a cliff that is 8.0 m above the water. One diver runs of

Alika [10]

Answer:

Number of revolutions=1.532 revolutions

Explanation:

Given data

Distance s=8.0 m

Angular speed a=1.2 rev/s

To find

Number of revolutions

Solution

From the equation of simple motion we not that

$S=ut+1/2gt^{2}\\ where\\u=0\\So\\8.0m=0+(1/2)(9.8m/s^{2} )t^{2}\\ t^{2}=\frac{8.0m}{0.5*9.8m/s^{2} } \\ t^{2}=1.63\\t=\sqrt{1.63} \\t=1.28s$

So for the number of revolutions she makes is given as

$n=a*t\\n=(1.2rev/s)(1.28s)\\n=1.532revolutions$

8 0

3 years ago

Look at the advertisement. A poster that reads, "There's more to resistors than resistance. If you're really serious about cost,

ozzi

Options:

universal appeal

flattery

association

bandwagon

Answer: Association

Explanation: Advertisement is a marketing technique through which business organisations utilize the opportunities provided by both the print, electronic and other channels of communication to market a product to the target audience.

Association Advertising is a type of Advertising where certain attributes which have been known to be associated with good and quality products are used to market the products to the target audience.

7 0

3 years ago

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