All categories

2 years ago

A microscope allowed Hooke to see "tiny rectangular rooms," which he called_____________?

Physics

2 answers:

-Dominant- [34]2 years ago

8 0

Cells
Hope this helps!!

Finger [1]2 years ago

4 0

they are called "cells"

hope this is the answer is what your looking for.

You might be interested in

A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul

Allisa [31]

Here in this case we can use work energy theorem

As per work energy theorem

Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

$KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2$

also we know that

$I = \frac{2}{5}mR^2$

$w= \frac{v}{R}$

Now kinetic energy is given by

$KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2$

$KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2$

$KE = \frac{7}{10}mR^2w^2$

$KE = \frac{7}{10}*150*(0.200)^2(50)^2$

$KE = 10500 J$

Now by work energy theorem

Work done = 10500 - 0 = 10500 J

So in the above case work done on sphere is 10500 J

7 0

2 years ago

A 300 g ball and a 600 g ball are connected by a 40-cm-lon massless, rigid rod. The structure rotates about its center of me at

Readme [11.4K]

Answer:

KE = 1.75 J

Explanation:

given,

mass of ball, m₁ = 300 g = 0.3 Kg

mass of ball 2, m₂ = 600 g = 0.6 Kg

length of the rod = 40 cm = 0.4 m

Angular speed = 100 rpm= $100\times \dfrac{2\pi}{60}$

=10.47\ rad/s

now, finding the position of center of mass of the system

r₁ + r₂ = 0.4 m.....(1)

equating momentum about center of mass

m₁r₁ = m₂ r₂

0.3 x r₁ = 0.6 r₂

r₁ = 2 r₂

Putting value in equation 1

2 r₂ + r₂ = 0.4

r₂ = 0.4/3

r₁ = 0.8/3

now, calculation of rotational energy

$KE = \dfrac{1}{2}I_1\omega^2+\dfrac{1}{2}I_2\omega^2$

$KE = \dfrac{1}{2}\omega^2 (I_1 +I_2)$

$KE = \dfrac{1}{2}\omega^2 (m_1r_1^2 +m_2r^2_2)$

$KE = \dfrac{1}{2}\times 10.47^2(0.3\times (0.8/3)^2 +0.6\times (0.4/3)^2)$

KE = 1.75 J

the rotational kinetic energy is equal to 1.75 J

7 0

3 years ago

What is the definition of electrical power?

Nonamiya [84]

An electric power measure the rate of electrical energy transfer by an electric circuit per unit of time.

6 0

1 year ago

a typical cmall flashlight contains two batteries each having na emf of 2.0 v connected in series with a bulb havin ga resistanc

Helen [10]

Answer:

P = 0.25 W

Explanation:

Given that,

The emf of the battry, E = 2 V

The resistance of a bulb, R = 16 ohms

We need to find the power delivered to the bulb. We know that, the formula for the power delivered is given by :

$P=\dfrac{V^2}{R}\\\\P=\dfrac{2^2}{16}\\\\=0.25\ W$

So, 0.25 W power is delivered to the bulb.

5 0

2 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

2 years ago