A 20kg bike accelerates at 10 m/s2. With what force was the person padeling?

Three light bulbs are connected to a battery in a series circuit.How will the bulbs behave if the circuit is closed

Drupady [299]

The answer is

<em>All the three light bulbs will glow</em>

hope this helps

6 0

3 years ago

Read 2 more answers

If a ball starts with the velocity of 2.0m/s and accelerates with a constant rate of 0.50m/s squared for 2.5s what is the balls

baherus [9]

Acceleration can be any change in speed, increasing or decreasing.
You haven't said whether the ball is speeding up or slowing down.

If its acceleration is positive ... speed is increasing ... then in 2.5 seconds,
it GAINS (0.5 m/s² x 2.5 sec) = 2.5 m/s of speed. Added to its initial
speed of 2.0 m/s, it ends up moving at 4.5 m/s.

If its acceleration is negative ... speed is decreasing ... then in 2.5 seconds,
it LOSES (0.5 m/s² x 2.5 sec) = 2.5 m/s of speed. Added to its initial
speed of 2.0 m/s, it ends up moving at -0.5 m/s. That means that it ends up
moving in the opposite direction compared to its direction at the beginning of
the change.

7 0

2 years ago

Use the work—energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resist

andreyandreev [35.5K]

Answer:

a) It is moving at $43.15\frac{m}{s^{2}}$ when reaches the ground.

b) It is moving at $101.44\frac{m}{s^{2}}$ when reaches the ground.

Explanation:

Work energy theorem states that the total work on a body is equal its change in kinetic energy, this is:

$W=K_f-K_i$ (1)

with W the total work, Ki the initial kinetic energy and Kf the final kinetic energy. Kinetic energy is defined as:

$K=\frac{mv^2}{2}$ (2)

with m the mass and v the velocity.

Using (2) on (1):

$W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$ (3)

In both cases the total work while the objects are in the air is the work gravity field does on them. Work is force times the displacement, so in our case is weight (w=mg) of the object times displacement (d):

$W=Fd=wd=mgd$ (4)

Using (4) on (3):

$mgd=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$ (5)

That's the equation we're going to use on a) and b).

a) Because the branch started form rest initial velocity (vi) is equal zero, using this and solving (5) for final velocity:

$v_f=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*95}$

$v_f=43.15\frac{m}{s^{2}}$

b) In this case the final velocity of the boulder is instantly zero when it reaches its maximum height, another important thing to note is that in this case work is negative because weight is opposing boulder movement, so we should use -mgd:

$-mgd=-\frac{mv_i^2}{2}$

Solving for initial velocity (when the boulder left the volcano):

$v_i=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*525}$

$v_i=101.44 \frac{m}{s^{2}}$

3 0

2 years ago

An 85 kg man and his 35 kg daughter are sitting on opposite ends of a 3.00 m see-saw. The see-saw is anchored in the center. If

wolverine [178]

Answer:

0.54m

Explanation:

Step one:

given data

length of seesaw= 3m

mass of man m1= 85kg

weight = mg

W1= 85*10= 850N

mass of daughter m2= 35kg

W2= 35*10= 350N

distance from the center= (1.5-0.2)= 1.3m

Step two:

we know that the sum of clockwise moment equals the anticlockwise moment

let the distance the must sit to balance the system be x

taking moment about the center of the system

350*1.3=850*x

455=850x

divide both sides by 850

x=455/850

x=0.54

Hence the man must sit 0.54m from the right to balance the system

3 0

3 years ago

Because the pressure falls, water boils at a lower temperature with increasing altitude. Consequently, cake mixes and boiled egg

AlexFokin [52]

Answer:

1) The boiling point of water reduces by 3.28°C at 1,000 m above sea-level

2) The boiling point of water reduces by 6.56°C at 2,000 m above sea-level

Explanation:

The variation of the boiling point of water with elevation is given as follows

The boiling point reduces by 0.5°C for every 152.4 meter increase in elevation

At sea-level, the boiling point temperature of water = 100°C

1) At 1,000 m elevation, the boiling point temperature, T = 100 - (1,000/152.4) × 0.5 ≈ 96.72 °C

Therefore, the boiling point of water reduces by 100° - 96.72° = 3.28°C at 1,000 m above sea-level

2) At 2,000 m elevation, the boiling point temperature, T = 100 - (2,000/152.4) × 0.5 ≈ 93.44°C

The boiling point of water reduces by 100° - 93.44° = 6.56°C at 2,000 m above sea-level

7 0

3 years ago