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jolli1 [7]
3 years ago
12

The distance xm travelled by a particle in time t seconds is described by the equation x = 10 + 12tsquare, Find the average spee

d of the particle between the time interval t = 2s and t = 5s
A. 60 ms-1
B. 72 ms-1
C. 84 ms-1
D. 108 ms-1​
Physics
1 answer:
Paraphin [41]3 years ago
5 0

After 2 seconds the particle will be in position

x(2)=10+12\cdot 2^2=10+12\cdot 4=10+48=58

After 5 seconds the particle will be in position

x(5)=10+12\cdot 5^2=10+12\cdot 25=10+300=310

So, the particle will travel

310-58=252

meters in 3 seconds, for an average speed of

\dfrac{252}{3}=84

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In need help I need someone that is really good at physics
Aloiza [94]
10/70×360°
=51.4°

hope thus helps
5 0
3 years ago
A paper clip that has a mass of 1.5 grams is thrown into the air and initially has a kinetic energy of 0.013 J.
Aleksandr [31]

answer

v = 4.2\frac{m}{s}

set up an equation

we can use the formula for kinetic energy since we know mass and kinetic energy

K = \frac{mv^{2}}{2}

v = \sqrt{\frac{2K}{m} }

values

K = 0.013 J

m = 1.5g = 0.0015 kg

plug in values

v = \sqrt{\frac{2*0.013}{0.0015} }

v = 4.2\frac{m}{s}

3 0
3 years ago
Una flecha tiene una rapidez de lanzamiento inicial de 18 m/s. Si debe dar en un blanco a 31 m de distancia, que está a la misma
maks197457 [2]

Answer:

34.8 and 55.2º

Explanation:

This is a projectile launching exercise, as we are told that the range of the arrow must be equal to its range and = 31 m let's use the equation

         

The scope equation is

         R = v₀² sin 2θ /g

         sin 2 θ = R g / v₀²

         sin 2 θ = 31 9.8 / 18²

         2 θ = sin⁻¹ 0.93765

          θ = 34.8º

At the launch of projectiles we have two complementary angles with the same range in this case 34.8 and (90-34.8) = 55.2º

4 0
3 years ago
An eagle is flying horizontally 16.4 meters above a lake at a speed of 9.3 m/s, carrying a small pumpkin in its talons. The pump
Dima020 [189]

Answer:

The horizontal distance the pumpkin will travel after it slips from the eagle is 17.02 m

Explanation:

Given;

height above the ground, h = 16.4 m

speed of the eagle, v = 9.3 m/s

The time it will take the pumpkin to fall at the given height is calculated as;

t = \sqrt{\frac{2h}{g} }\\\\t =  \sqrt{\frac{2*16.4}{9.8} }\\\\t = 1.83 \ s

The horizontal distance traveled at this time is given by;

x = vt

x = (9.3)(1.83)

x = 17.02 m

Therefore, the horizontal distance the pumpkin will travel after it slips from the eagle is 17.02 m

4 0
2 years ago
Which mineral is hard enough to scratch calcite but is not hard enough to scratch amphibole
Salsk061 [2.6K]

Answer:

Muscovite mica. Flourite. Olivine.

8 0
2 years ago
Read 2 more answers
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