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murzikaleks [220]

3 years ago

13

A 2,000-kg test car, traveling 60 m/s hits a brick wall. Using motion pictures, the time involved is determined to be 0.050 s. W

hat is the change in momentum of the car?
A) +1.2 x 10^5

B)-1.2 x 10^5

1 answer:

Naily [24]3 years ago

8 0

change in momentum of car is given as

$\Delta P = P_f - P_i$

here we have

$\Delta P = m(v_f - v_i)$

given that

m = 2000 kg

$v_f = 0$

$v_i = 60 m/s$

now we have

$\Delta P = 2000(0 - 60)$

$\Delta P = - 1.2 \times 10^5 kg m/s$

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If a spear is jabbeb exactly at that place where the fish appears in water, the fish is not killed

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3 years ago

A proton is released in a uniform electric field, and it experiences an electric force of 2.04×10−14 N toward the south.

Lemur [1.5K]

Answer:

Part A:

$E=127500N/C\\E=1.275*10^{5} N/C$

Part B:

Option B (Towards the South)

Explanation:

Part A:

Magnitude if electric field E:

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Part B:

Option B (Towards the South)

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3 years ago

What effect does Earth’s spherical shape have on the amount of sunlight that reaches the equator?

Natasha_Volkova [10]

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3 years ago

Read 2 more answers

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

3 years ago

What is the mechanical advantage of a nail puller where you exert a force 45 cm from the pivot and the nail is 1.8 cm on the oth

Mandarinka [93]

Answer:

Explanation:

Mechanical Advantage is the ratio of the distance of the input load (Li)from the pivot to the output load applied to the pivot(Lo)

MA = Li/Le

Given;

Li = 45cm

Lo = 1.8cm

MA = 45/1.8

MA = 25

Hence the mechanical advantage is 25

Also MA is expressed in terms of the force ratio which is the ratio of the Load to the effort applied.

MA = Load/Effort

Given

Load = 1250N

MA = 25

Effort = ?

Substitute

25 = 1250/Effort

Effort = 1250/25

Effort = 50N

Hence the minimum force exerted on the load is 50N

3 0

3 years ago