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3 years ago

7

50 N of force is exerted on a crate. It accelerates at a rate of 1.2 m/s/s. Calculate its mass

1 answer:

vitfil [10]3 years ago

4 0

Answer:

m = F/a = 50 / 1.2 = 41.6666.... 42 kg

Explanation:

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Margarita [4]

1) The object slows down due to kinetic friction.
2) The coefficient of kinetic friction is higher on a carpet than on the bare floor, therefore the object would slow down quicker on the carpet

7 0

2 years ago

Read 2 more answers

Let's examine over the next few questions an NFL kick as described in the 3rd Law video. The announcer claimed that the ball is

hichkok12 [17]

Answer:

8451.62109367671 Newtons

Explanation:

$1\ kg=1\times 9.8066500286389=9.8066500286389\ N$

$1 \lb=2.2046226218488\ kg$

It is known that

$1\ lbs=\dfrac{9.8066500286389}{2.2046226218488}=4.4482216282508\ N$

So,

$1900\ lbs=1900\times 4.4482216282508\\\Rightarrow 1900\ lbs=8451.6210936765\ N$

The force in Newtons is 8451.6210936765 Newtons

4 0

3 years ago

A bicyclist is traveling at +25m/s when he begins to decelerate at -4m/s2. How fast is he traveling after 5 seconds

kotegsom [21]

Answer:

+5m/s

Explanation:

When doing the math we figure out that e is going to be slowing down at -4m/s² for 5 seconds. In total he is slowing down -20m/s which we take from the total speed of +25m/s to get his current new speed.

7 0

2 years ago

A paperweight consists of a 8.55-cm-thick plastic cube. Within the plastic a thin sheet of paper is embedded, parallel to opposi

ella [17]

Answer:

$\mu=1.5322$

Explanation:

Given:

Thickness of the paperweight cube, $x=8.55\ cm$

apparent depth from one side of the inbuilt paper in the plastic cube, $i=4.02\ cm$

apparent depth from the other side of the inbuilt paper in the plastic cube, $i'=1.56\ cm$

Now as we know that refractive index is given as:

$\rm \mu=\frac{real\ depth}{apparent\ depth}$

Let the real depth form first side of the slab be, $d$

Then the depth from the second side of the slab will be, $d'=x-d=8.55-d$

Since refractive index for an amorphous solid is an isotropic quantity so it remains same in all the direction for this plastic.

$\mu=\mu'$

$\frac{d}{i} =\frac{d'}{i'}$

$\frac{d}{4.02}=\frac{8.55-d}{1.56}$

$d=6.1597\ cm$

Now the refractive index:

$\mu=\frac{d}{i}$

$\mu=\frac{6.1596}{4.02}$

$\mu=1.5322$

7 0

2 years ago

6. A .25 kg arrow with a velocity of 12 m/s to the west strikes and pierces the center of a 6.8 kg target. a. What is the final

Alenkasestr [34]

Answer:

(a) the final velocity of the combined mass is 9.43 m/s

(b) the decrease in kinetic energy during the collision is 386.1 J

Explanation:

Given;

mass of arrow, m₁ = 25 kg

initial velocity of arrow, u₁ = 12 m/s

mass of target, m₂ = 6.8 kg

initial velocity of the target, u₂ = 0

Part (a)

From the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = v(m₁+m₂)

where;

v is the final velocity of the combined mass

25 x 12 + 0 = v(25 + 6.8)

300 = v(31.8)

v = 300/31.8

v = 9.43 m/s

Part(b)

Kinetic Energy, K.E = ¹/₂mv²

Initial kinetic energy = ¹/₂m₁u₁² + ¹/₂m₂u₂² = ¹/₂ x 25 x (12)² + 0 = 1800 J

Final kinetic energy = ¹/₂m₁v² + ¹/₂m₂v² = ¹/₂v²(m₁ + m₂)

= ¹/₂ x (9.43)²(25+6.8)

= 1413.91 J

Decrease in kinetic energy = Initial K.E - Final K.E

Decrease in kinetic energy = 1800J - 1413.91 J = 386.1 J

4 0

3 years ago