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3 years ago

What is the potential energy of a puppy that weighs 18 N istting in a high chair 2 m high?

Physics

1 answer:

kykrilka [37]3 years ago

6 0

Answer:

Potential energy =mass* acceleration due to gravity * height

mass*acceleration due to gravity =weight

hence potential energy of the puppy= weight * height

=18*2

=36 joule

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A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete

Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is $6.25\times 10^{- 3} ft/s^{2}$

Solution:

As per the question:

Constant velocity of the horse in the horizontal, $v_{x} = 1 ft/s$

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

$20^{2} + 10^{2} = l^{2}$

$l^{2}= 500$

Now,

$x^{2} + y^{2} = 500$ (1)

Differentiating equation (1) w.r.t 't':

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$x\frac{dx}{dt} = - y\frac{dy}{dt}$

where

$\frac{dx}{dt}$ = Rate of change of displacement along the horizontal

$\frac{dy}{dt}$ = Rate of change of displacement along the vertical

$v_{x}$ = velocity along the x-axis.

$v_{y}$ = velocity along the y-axis

$xv_{x} = -yv_{y}$

$v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s$

$|v_{y}| = 0.5\ ft/s$

Acceleration of the hay bale is given by the kinematic equation:

$v_{y}^{2} = u_{y} + 2ay$

$(-0.5)^{2} =0 + 2ay$

$0.25 = 2ay$

$\frac{0.25}{2y} = a$

$a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}$

7 0

4 years ago

One's perception depends on the way the brain interprets the stimuli. true or false

bulgar [2K]

The answer is true .

5 0

4 years ago

Which of the following is true regarding momentum?

kakasveta [241]

Increasing the mass of an object increases its momentum.

Explanation:

Momentum of an object is measured as the quantity of motion done by the object.
It is calculated using the formula, p = m × v where m is mass of the object and v is the velocity of the object.
As momentum and mass vary proportionally, as seen in the formula, increasing the mass of an object will also increase its momentum.

8 0

3 years ago

Two pro baseball players of the same height are testing how far they can throw. One players throws the ball at a 40.0 degrees an

Viktor [21]

Answer:

The distance traveled by the ball before it lands in the other player's glove is 130.96 m.

Explanation:

Given;

angle of projection of the ball, θ = 40⁰

initial velocity of the ball, u = 36.1 m/s

The distance traveled by the ball before it lands in the other player's glove is the range of the projectile, calculated as follows;

$R = \frac{u^2 sin(2\theta)}{g} \\\\R= \frac{36.1^2 \times sin(2\times 40)}{9.8} \\\\R = \frac{36.1^2 \times sin(80)}{9.8} \\\\R = 130.96 \ m$

Therefore, the distance traveled by the ball before it lands in the other player's glove is 130.96 m.

6 0

3 years ago

The Achilles tendon connects the muscles in your calf to the back of your foot. When you are sprinting, your Achilles tendon alt

lesantik [10]

Answer:

(a) $\triangle l=5 mm$

(b) 0.033

Explanation:

(a)

Force F=mg where m is mass and g is acceleration due to gravity whose value is taken as $9.81 m/s^{2}$

However, for this case, the maximum force is 8 times the weight of runner hence F=8mg

Assume Young's modulus for tendon is $0.15*10^{10} N/m^{2}$

Young's modulus is given by

$E=\frac {Fl}{A\triangle l}$ and $\triangle l=\frac {Fl}{EA}$ and substituting F with 8mg we obtain $\triangle l=\frac {8mgl}{EA}$

Where E is young's modulus, l is stretched length and \triangle l is change in length

Substituting m as 70 kg, g as $9.81 m/s^{2}$ , l as 15cm=0.15 m, E as $0.15*10^{10} N/m^{2}$ and A as $110 m^{2}=0.000110 m^{2}$

$\triangle l=\frac {8*70 Kg*9.81 m/s^{2}*0.15m}{0.15*10^{10} N/m^{2} *0.00011 m^{2}}=0.004994182 m$

$\triangle l=5 mm$

(b)

Strain, $\epsilon=\frac {\triangle l}{l}$ and the fraction of tendon’s length is the ratio of change in length to the stretched length

The fraction of tendon, f is given by

$f=\frac {\triangle l}{l}$ . Substituting $\triangle l$ with 0.005m and l with 0.15m we obtain

$\epsilon=f=\frac {0.005}{0.15}=\frac {1}{30}=0.033$

Therefore, fraction of the tendon’s length is 0.033

5 0

4 years ago