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Nina [5.8K]
2 years ago
5

A 15kg penguin slides on its belly down an icy, frictionless glacier. If the penguin started from rest and reaches a speed of 11

m/S when at the bottom, how high is the glacier? (Metres)
Physics
1 answer:
Schach [20]2 years ago
8 0

Answer:

Mechanical Energy (initial) = Mechanical energy (final)

Ep(initial) + Ek(initial) = Ep(final) + Ek(final)

mgh(initial) + 1/2mv²(initial) = mgh(final) + 1/2mv²(final)

m being the mass of the penguin (kg)

g being gravitational acceleration (9.80 m/s²)

v(initial) being zero

h(final) being zero,i.e. final height is zero

v(final) being final velocity

solve for h(initial) being initial height at the top of the glacier

answer will be in metres

NB: don't forget to square the velocity

Explanation:

Since they mentioned frictionless, we know that this is a closed system therefore we are free to use this equation of conservation of mechanical energy

Don't forget to rate answer

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An object travels along a right triangle with sides 3 m, 4 m, and 5 m. It returns back to the initial position.
stiv31 [10]
When it travels 3m ,4m and 5m it means 12m is right answer.
3 0
3 years ago
A boat takes off from a dock at 2.5 m/s and speeds up at 4.2 m/s squared for six seconds how far has the most traveled
GaryK [48]

The boat's position x relative to its starting point x_0=0 is determined by

x=x_0+v_0t+\dfrac12at^2

where v_0 is its initial velocity, a is its acceleration, and t is time. After t=6\,\mathrm s, the boat has traveled

x=\left(2.5\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(4.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2

\implies x=91\,\mathrm m

3 0
3 years ago
The ideal mechanical advantage of an inclined plane is 3.5, and its efficiency is 0.6. What is the mechanical advantage of the i
Nadya [2.5K]
Mechanical advantage = ideal mechanical advantage x efficiency = 3.5 x 0.6 = 2.1
The mechanical advantage of the inclined plane is 2.1

7 0
3 years ago
Bob wanted to investigate the effects of plant fertilizer. He bought three identical plants and gave plant 1 fertilizer every mo
olga_2 [115]

Answer:

TIME he applied the fertilizer to each plant

Explanation:

Independent variable in an experiment is the variable that is subject to change or manipulation by the experimenter. In this experiment, Bob wanted to investigate the effects of plant fertilizer. Bob sets up the experiment by applying the fertilizer to each plant at DIFFERENT TIMES i.e. plant 1-every morning, plant 2-once a week, plant 3-never.

Based on this, it is obvious that the independent or manipulated variable is the TIME at which he applied the fertilizer. On the other hand, the dependent or measured variable is the height of the plants.

8 0
2 years ago
The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

6 0
3 years ago
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