Ask question

Ask question

All categories

2 years ago

15

Suppose you have two cars, and the larger one is twice as massive as the smaller one. If you and a friend push on them so that t

heir acceleration are equal, how must the forces applied to the cars compared?

1 answer:

dexar [7]2 years ago

3 0

The force on the smaller car will be less thab the large one because Force is directly proportional to mass

You might be interested in

The earth is made mostly of metals and rocks. Where did this material come from? It was made by nuclear fission of uranium and o

Olegator [25]

This is very very cool!

5 0

3 years ago

Which organism in the food web above is sometimes a first-level consumer and sometimes a second-level consumer?Explain

scoray [572]

The answer is: Mouse/Herbivore.

First organism is always a producer (plant) such as grass.
The 2nd organism is the first level or primary consumer. Ex. Mouse - It eats the producer, so it is a herbivore.

4 0

3 years ago

Which of the following are single-displacement reactions?

ololo11 [35]

Answer:

A & D

Explanation:

A single-displacement reaction is a chemical reaction whereby one element is substituted for another one in a compound and thereby generating a new element and also a new compound as products.

From the options, only options A & D fits this definition of single-displacement reactions.

For option D: Both left and hand and right hand sides each have one element and one compound. We can see that K is substituted from KBr to join Cl to form KCl and Br2 on the right hand side.

For option A: Both left and hand and right hand sides each have one element and one compound. We can see that OH is substituted from 2H2O to join Mg to form Mg(OH)2 and H2 on the right hand side.

The other options are not correct because they don't involve only and element and a compound on each side of the reaction.

6 0

2 years ago

Read 2 more answers

Two planets have masses 2 x 10^23 kg and 5 x 10^22 kg, and the distance between them is 3 x 10^16 m. Calculate the gravitational

xxTIMURxx [149]

Explanation:

given,

mass of one planet (m1)=2*10^23 kg

mass of another planet (m2)=5*10^22kg

distance between them(d)=3*10^16m

gravitational constant(G)=6.67*10^-11Nm^2kg^-2

gravitational force between them(F)=?

we know,

F=Gm1m2/d^2

or, F=6.67*10^-11*2*10^23*5*10^22/(3*10^16)^2

or, F=6.67*2*5*10^-11+23+22/3*3*10^32

or, F=66.7*10^34/9*10^32

or, F=7.41*10^34-32

•°• F=7.41*10^2

thus, the gravitational force between them is 7.14*10^2

3 0

2 years ago

Recall from Chapter 1 that a watt is a unit of en- ergy per unit time, and one watt (W) is equal to one joule per second ( J·s–1

harkovskaia [24]

Answer:

Explanation:

The energy of a photon is given by the equation $E_p=h f$ , where h is the <em>Planck constant</em> and f the frequency of the photon. Thus, N photons of frequency f will give an energy of $E_N=N h f$ .

We also know that frequency and wavelength are related by $f=\frac{c}{\lambda}$ , so we have $E_N=\frac{N h c}{\lambda}$ , where c is the <em>speed of light</em>.

We will want the number of photons, so we can write

$N=\frac{\lambda E_N}{h c}$

We need to know then how much energy do we have to calculate N. The equation of power is $P=E/t$ , so for the power we have and considering 1 second we can calculate the total energy, and then only consider the 4% of it which will produce light, or better said, the N photons, which means it will be $E_N$ .

Putting this paragraph in equations:

$E_N=(\frac{4}{100})E=0.04Pt=(0.04)(100W)(1s)=4J$ .

And then we can substitute everything in our equation for number of photons, in S.I. and getting the values of constants from tables:

$N=\frac{\lambda E_N}{h c}=\frac{(520 \times10^{-9}m) (4J)}{(6.626\times10^{-34}Js) (299792458m/s)}=1.047 \times10^{19}$

3 0

2 years ago