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3 years ago

Q1.

Physics

1 answer:

Vesnalui [34]3 years ago

4 0

Answer:

P = 450 J

Explanation:

Given that,

Mass of a child, m = 18 kg

The vertical distance from the top to the bottom of the slide is 2.5 metres.

The Gravitational field strength = 10 N/kg

We need to find the decrease in gravitational potential energy of the child sliding from the top to the bottom of the slide.

The formula for the gravitational potential energy is given by :

P = mgh

Substituting all the values,

P = 18 kg × 10 m/s² × 2.5 m

P = 450 J

Hence, the decrease in gravitational potential energy is 450 J.

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3 years ago

An airplane takes off a runway at a constant speed of 49m/s at constant angle 30 to the horizontal

joja [24]

Complete Question

An airplane takes off a runway at a constant speed of 49 m/s at constant angle 30 to the horizontal.How high (in meters ) is the airplane above the ground 13 seconds after takeoff?

Answer:

The height is $H = 318.5 \ m$

Explanation:

From the question we are told that

The speed at which the plane takes off is $u = 49 \ m/s$

The angle at which it takes off is $\theta = 30 ^o$

The time taken is $t = 13 s$

The vertical distance traveled is mathematically represented as

$H = u sin \theta t$

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6 0

3 years ago

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

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3 years ago