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marta [7]
2 years ago
13

What do you think Jose’s grandmother meant by saying that the bread was too dense? What other things can you think of that are “

dense”?
Physics
1 answer:
Deffense [45]2 years ago
8 0
The bread was most likely heavy
You might be interested in
Calculate the kinetic energy of Julie, a 60 kg biker, traveling at a velocity of 8 m/s to the right
Vadim26 [7]

Answer:

1,920 Joules

Explanation:

K.E. = 1/2  mv2

so  K.E. =  1/2 (60)(8x8) = 1,920 Joules

8 0
2 years ago
If you travel 1.7 km north from your house at noon, and at 6:00 PM you travel 5.4 km south, what is your displacement? 3.7 km no
Yakvenalex [24]

<u>Answer</u>

3.7 Km south


<u>Explanation</u>

The definition of displacement is the distance traveled in a specific direction. It is the vector quantity. We add displacements like the way we add vectors.

Taking the direction towards North to be positive (+1.7 Km), the distance towards south would be negative (-5.4 Km).

Now lets add the two values.

(+1.7) + (-5.4) = 1.7 - 5.4

                    = - 3.7 Km      But negative was towards south.

∴ Answer = 3.7 Km south.


6 0
3 years ago
A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the
BartSMP [9]

Answer:

r=1.14m

Explanation:

\theta is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

F_m=F_c\\qvB=F_c(1)

F_c is the centripetal force and is defined as:

F_c=m\frac{v^2}{r}

Here v is the proton's speed and r is the radius of the circular motion. Replacing this in (1) and solving for r:

qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}

Recall that 1 J is equal to 6.242*10^{12}MeV, so:

4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J

We can calculate v from the kinetic energy of the proton:

K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}

Finally, we calculate the radius of the proton path:

r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m

8 0
3 years ago
How will you join three resistances, each of 2 ohm so that the effective resistance is 3 ohm?
svet-max [94.6K]
First I will parallel two of the resistors, creating a net 1 ohm. Then I will series that with the remaining 2-ohm resistor, resulting in 3 ohms.
5 0
3 years ago
Read this excerpt from Through the Looking-Glass by Lewis Carroll.
Vanyuwa [196]

Answer:

Why does Alice forget the name of the woods and her own name?

6 0
3 years ago
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