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3 years ago

What is the student's kinetic energy at the bottom of the hill if he is moving

Physics

1 answer:

soldi70 [24.7K]3 years ago

5 0

Answer:

KE = 10530 J or 10.53 KJ

Explanation:

The formula for kinetic energy is KE = 1/2 mv^2

Let's apply the formula:

KE = 1/2 mv^2

KE = 1/2 (65kg) (18m/s)^2

KE = 10530 J or 10.53 KJ

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Distance (m)

vitfil [10]

Answer:41 m/s

Explanation:

3 0

3 years ago

An automobile can be considered to be mounted on four identical springs as far as vertical oscillations are concerned. The sprin

iogann1982 [59]

Answer:

Therefore, the spring constant of each spring = 1.6 × 10⁻⁶ kg/s².

Explanation:

The period (T) of a spring in oscillation = 2π √(m/k)............. equation 1

Where m = mass acting on the spring (kg), k = spring constant of the spring (kg/s²).

Making k the subject of equation 1

k = T²/(4π²×m) .......................... equation 2

From the question, F = 4.42 Hz,

since T = 1/F

then, T = 1/F = 1/4.42 =0.226 s, π = 3.143

since the weight of the mass is evenly distributed over the four identical spring, Hence

m = 1450/4 = 362.5 kg

Substituting these values into equation 2

k = 0.226/{(4×3.143²)362.5}

k = 0.226/(14323.751)

k = 0.0000016 kg/s²

k = 1.6 × 10⁻⁶ kg/s².

Therefore, the spring constant of each spring = 1.6 × 10⁻⁶ kg/s².

5 0

3 years ago

Review 1: A plane is located x = 40 miles (horizontally) away from an airport at an altitude of h miles. Radar at the airport de

-Dominant- [34]

Explanation:

Let h is the height of the plane above ground. x is the horizontal distance between the ground and the airport. Let s(t) is the distance between the plane and the airport. So,

$s(t)^2={h^2+x^2}$ ...........(1)

Given, h = 4, x = 40 and s(t) = -20 mph

Differentiate equation (1) wrt t

$2s(t)s'(t)=2x(t)x'(t)$

$x'(t)=\dfrac{s(t)s'(t)}{x(t)}$

When x = 40, $s(t)=\sqrt{40^2+4^2}=40.19\ m$

$x'(t)=\dfrac{-240s(t)}{x(t)}$

$x'(t)=\dfrac{-240\times 40.19}{40}$

$x'(t)=-241.14\ m/s$

So, the speed of the airplane is 241.14 m/s. Hence, this is the required solution.

8 0

3 years ago

A ball of plasticine is released from rest at height of 2.2 m above the ground. After touching the ground, the plasticine ball c

Anna35 [415]

The magnitude of the acceleration of the ball while coming to rest is 477.43 m/s²

The direction of the acceleration of the ball is downwards

The given parameters

initial velocity of the ball, u = 0

height above the ground, h = 2.2 m

time of motion of the ball, t = 96 ms = 0.096 s

The magnitude of the acceleration of the ball while coming to rest is calculated as;

let the downwards direction of the acceleration be positive

$h = ut + 0.5 at^2\\\\h = 0 + 0.5at^2\\\\h = 0.5 at^2\\\\a = \frac{h}{0.5t^2} \\\\a = \frac{2.2}{0.5 \times 0.096^2} \\\\a = 477.43 \ m/s^2$

The direction of the acceleration of the ball is downwards

Learn more here: brainly.com/question/15407740

4 0

2 years ago

The cornea behaves as a thin lens of focal lengthapproximately 1.80 {\rm cm}, although this varies a bit. The material of whichi

Keith_Richards [23]

Answer:

Explanation:

a )

from lens makers formula

$\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})$

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face

putting the values

$\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})$

1.462 = 2 - 1 / r₂

1 / r₂ = .538

r₂ = 1.86 cm .

= 18.6 mm .

b )

object distance u = 25 cm

focal length of convex lens f = 1.8 cm

image distance v = ?

lens formula

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}$

$\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}$

.5555 - .04

= .515

v = 1.94 cm

c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

= .776 mm

It will be a real image and it will be inverted.

5 0

3 years ago