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Gwar [14]
3 years ago
9

Please helppppppppppppp

Physics
1 answer:
miskamm [114]3 years ago
8 0

  • Force, F = 15 N
  • Mass,m = 3 kg
  • Time taken, t = 5 sec
  • (v-u) =?

<u>As we know </u>–

\qquad\purple{\bf  \twoheadrightarrow  F = m a }

\qquad\bf \twoheadrightarrow  F = m \dfrac{ v-u}{t}

\qquad\sf  \twoheadrightarrow 15 = 3 \times  \dfrac{v-u}{5}

\qquad\sf  \twoheadrightarrow  3 \times  \dfrac{v-u}{5} = 15

\qquad\sf  \twoheadrightarrow  \dfrac{3}{5} \times (v-u) = 15

\qquad\sf  \twoheadrightarrow (v-u)= 15 \times \dfrac{5}{3}

\qquad\sf  \twoheadrightarrow  (v-u)= \cancel{15} \times \dfrac{5}{\cancel{3}}

\qquad\sf  \twoheadrightarrow  v-u = 5 \times 5

\qquad\purple{\bf  \twoheadrightarrow  v-u = 25 }

  • Henceforth, the change in the speed of the body is 25 m/s.
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The work done to compress a spring with a force constant of 290.0 N/m a total of 12.3 mm is: a) 3.57 J b) 1.78 J c) 0.0219 J d)
iren2701 [21]

Answer:

Work done, W = 0.0219 J

Explanation:

Given that,

Force constant of the spring, k = 290 N/m

Compression in the spring, x = 12.3 mm = 0.0123 m

We need to find the work done to compress a spring. The work done in this way is given by :

W=\dfrac{1}{2}kx^2

W=\dfrac{1}{2}\times 290\times (0.0123)^2

W = 0.0219 J

So, the work done by the spring is 0.0219 joules. Hence, this is the required solution.

7 0
4 years ago
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luda_lava [24]

Answer:

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Explanation:

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5 0
3 years ago
A wheel 1.0 m in radius rotates with an angular acceleration of 4.0rad/s2 . (a) If the wheel’s initial angular velocity is 2.0 r
Oliga [24]

Answer:

(a) ωf= 42 rad/s

(b) θ = 220 rad

(c) at = 4 m/s²  ,  v = 42 m/s

Explanation:

The uniformly accelerated circular movement,  is a circular path movement in which the angular acceleration is constant.

There is tangential acceleration (at ) and is constant.

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t  Formula (1)

θ=  ω₀*t + (1/2)*α*t²  Formula (2)

at = α*R  Formula (3)

v= ω*R  Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular velocity ( rad/s)

ωf: final angular velocity ( rad/s)

R : radius of the circular path (cm)

at : tangential acceleration (m/s²)

v : tangential speed (m/s)

Data

α = 4.0 rad/s² : wheel’s angular acceleration

t = 10 s

ω₀ = 2.0 rad/s  : wheel’s initial angular velocity

R = 1.0 m  : wheel’s radium

(a)  Wheel’s angular velocity after 10 s

We replace data in the formula (1):

ωf= ω₀ + α*t

ωf= 2 + (4)*(10)

ωf= 42 rad/s

(b) Angle that rotates the wheel in the 10 s interval

We replace data in the formula (2):

θ=  ω₀*t + (1/2)*α*t²

θ=  (2)*(10) + (1/2)*(4)*(10)²

θ=  220 rad  

θ=  220 rad  

(c) Tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval

We replace data in the Formula (3)

at = α*R = (4)(1)

at = 4 m/s²

We replace data in the Formula (4)

v= ω*R = (42)*(1)

v = 42 m/s

6 0
4 years ago
1, 2, &amp; 3.........................
oksian1 [2.3K]

Answer:

1 is correct

2 is -150  

4 0
3 years ago
If the sun is 400 times bigger than the moon, how couild the moon possibly cover the sun during a solar eclipse?​
AlekseyPX

Explanation:

the Moon passes between Earth and the Sun Even though the Moon is much smaller than the Sun, because it is just the right distance away from Earth, the Moon can fully block the Sun's light from Earth's perspective This completely blocks out the Sun's light

3 0
3 years ago
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