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kondaur [170]
2 years ago
13

FASTTT FOR BRAINLIST

Physics
1 answer:
telo118 [61]2 years ago
7 0
The answer should be A?
You might be interested in
6. During an impact time casting 5 x 10-45 a gulf club exerts an average impar
Novay_Z [31]

Answer:

2.5 × 10-⁴¹ Ns

Explanation:

Impulse

I = F × t

I = 5000 N × 5 × 10-⁴⁵ s

I = 25 × 10-⁴² Ns

I = 2.5 × 10-⁴¹ Ns

#LearnWithEXO

6 0
3 years ago
Consider three scenarios in which a particular box moves downward under the pull of gravity :
luda_lava [24]

Answer:

1. True  WA > WB > WC

Explanation:

In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.

A) Work is the product of force by distance and the cosine of the angle between them

    WA = W h cos 0

   WA = mg h

B) On a ramp without rubbing

     Sin30 = h / L

     L = h / sin 30

     WB = F d cos θ  

     WB = F L cos 30

     WB = mf (h / sin30) cos 30

     WB = mg h ctan 30

C) Ramp with rubbing

    W sin 30 - fr = ma

   N- Wcos30 = 0

   W sin 30 - μ W cos 30 = ma

    F = W (sin30 - μ cos30)

   WC = mg (sin30 - μ cos30) h / sin30

   Wc = mg (1 - μ ctan30) h

When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum

Let's review the claims

1. True The work of gravity is the greatest and the work where there is friction is the least

2 False. The job where there is friction is the least

3 False work with rubbing is the least

4 False work with rubbing is the least

5 0
3 years ago
Suppose that you are headed toward a plateau 55 meters high. If the angle of elevation to the top of the plateau is 40degrees​,
Iteru [2.4K]

Answer:

x=65.55m

Explanation:

Let x be the distance to the shore

From trigonometry properties:

tan(40^{o} )=\frac{55m}{x} \\x=\frac{55m}{tan(40^{o} )} \\x=65.55m

3 0
3 years ago
A jet plane is launched from a catapult on an aircraft carrier. In 2.0 s it reaches a speed of 42 m/s at the end of the catapult
Dvinal [7]

Answer:

Explanation:

Given

time taken t=2\ s

Speed acquired in 2 sec v=42\ m/s

Here initial velocity is zero u=0

acceleration is the rate of change of velocity in a given time

a=\frac{v-u}{t}

a=\frac{42-0}{2}=21\ m/s^2

Distance travel in this time

s=ut+0.5at^2

where  

s=displacement

u=initial velocity

a=acceleration

t=time

s=0+\0.5\times 21\times (2)^2

s=42\ m

so Jet Plane travels a distance of 42 m in 2 s                                

5 0
3 years ago
/Q: DHSMV definition/ i have to write 20 characters for whatever reason lol
cluponka [151]

<span><span>Department of Highway Safety and Motor Vehicles          OR</span></span>

<span><span /><span><span>Division of Highway Safety and Motor Vehicles </span></span></span>

6 0
3 years ago
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