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3 years ago

Second and third class levers are differentiated by __________.

Physics

2 answers:

ira [324]3 years ago

5 0

Answer:

i think it is b

hope this helps

Explanation:

azamat3 years ago

3 0

Answer: B

Explanation:the load in the second class lever is found in between the effort and the fulcrum

While in third class lever effort is found in between load and fulcrum

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A scientist measures the growth of a bamboo plant over time. The table above shows the results. What is the best interference fo

lisabon 2012 [21]

(B) 2.25cm

Explanation:

Given:

At 40 hours, the height of the bamboo plant is 2.1cm

At 50 hours, the height of the bamboo plant is 2.4cm

Height of the bamboo plant after 45 hours = ?

The difference in length from 40 to 50 hours = 2.4 - 2.1cm

= 0.3 cm

Mean of 40 and 50 is 45.

Thus,

At 45 hours, the height will increase by 0.3/2

= 0.15 cm

Height at 45 hour = 2.1 + 0.15cm

= 2.25cm

Therefore, the height of the plant after 45 hours is 2.25cm

7 0

3 years ago

A student conducts an experiment in which a cart is pulled by a variable applied force during a 2sec time interval. In trial 1,

Harlamova29_29 [7]

Answer:

change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

Explanation:

Let's look for the speed of the car

F = m a

a = F / m

We use kinematics to find lips

v = v₀ + a t

v = v₀ + (F / m) t

The moment is defined by

p = m v

The moment change

Δp = m v - m v₀

Let's replace the speeds in this equation

Δp = m (v₀ + F / m t) - m v₀

Δp = m v₀ + F t - m v₀

Δp = F t

We see that the change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

8 0

3 years ago

8. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bott

nasty-shy [4]

Answer: The ball is travelling with a speed of 5.5 m/s after hitting the bottle.

Explanation:

To calculate the speed of ball after the collision, we use the equation of law of conservation of momentum, which is given by:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

$m_1,u_1\text{ and }v_1$ are the mass, initial velocity and final velocity of ball.

$m_2,u_2\text{ and }v_2$ are the mass, initial velocity and final velocity of bottle.

We are given:

$m_1=0.4kg\\u_1=18m/s\\v_1=?m/s\\m_2=0.2kg\\u_2=0m/s\\v_2=25m/s$

Putting values in above equation, we get:

$(0.4\times 18)+(0.2\times 0)=(0.4\times v_1)+(0.2\times 25)\\\\v_1=5.5m/s$

Hence, the ball is travelling with a speed of 5.5 m/s after hitting the bottle.

5 0

3 years ago

Model rocket A is launched at an angle of 70° above the x axis, with an initial velocity of 40 m/s. An identical rocket B is lau

amid [387]

A. Rocket A will travel farther horizontally than rocket B.

This is because from the x axis, 40 m/s at 90 degrees travels directly vertical. 40 m/s at 70 degrees is slightly horizontal, so it will travel further horizontally.

3 0

3 years ago

When responding to sound, the human eardrum vibrates about its equilibrium position. Suppose an eardrum is vibrating with an amp

Paraphin [41]

Answer:

796.18 Hz

Explanation:

Applying,

Maximum velocity = Amplitude×Angular velocity

Therefore,

V' = A(2πf)............... Equation 1

Where V' = maximum velocity of the eardrum, A = Amplitude of vibration of the eardrum, f = frequency of the eardrum vibration, π = pie

make f the subject of the equation

f = V'/2πA................ Equation 2

From the question,

Given: V' = 3.6×10⁻³ m/s, A' = 7.2×10⁻⁷ m,

Constant: 3.14.

Substitute these values into equation 2

f = 3.6×10⁻³/( 7.2×10⁻⁷×2×3.14)

f = 796.18 Hz

6 0

3 years ago