Help please I’ll mark as brainliest

Lauren is comparing rocks from two different locations. She has the following

telo118 [61]

Answer:

the same type because they both formed from rock material.

Explanation:

4 0

3 years ago

you realize that your friend is clearly distracted and not paying attention to everything that is happening on the road. describ

alekssr [168]

A. ask him/her to pull over and offer to drive (if you have your license.)
B. ask him/her to call someone to come pick you guys up.
C. call a cab.

8 0

3 years ago

If an electron moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what are the speed of the electron and the

kodGreya [7K]

Answer:

a)

$v = 4.048 *10^6$ m/s

b)

Angular frequency = $1.92 * 10^7$

Explanation:

As we know

$v = \frac{qBr}{m}$

q is the charge on the electron = $3.2 * 10^{-19}$ C

B is the magnetic field in Tesla = $0.4$ T

r is the radius of the circle = $0.21$ m

mass of the electrons = $6.64 * 10^{-27}$ Kg

a)

Substituting the given values in above equation, we get -

$v = \frac{3.2 * 10^{-19}*0.4*0.21}{6.64 * 10^{-27}} \\v = 4.048 *10^6$ m/s

b)

Angular frequency =

$\frac{4.048 * 10^6 }{0.21} \\1.92 * 10^7$

8 0

3 years ago

An imaginary dense star z has 1/5 the radius of earth, but 1000 times the earths. how much would a mass weighing 1.0n on earth w

andrew11 [14]

A 1-newton mass on earth would be 1000 newtons on star Z.

Funny enough, stars like this exist~! They're called "Neutron Stars."

8 0

3 years ago

a boy pulls 23 kg box with a 100 N force at 39 above a horizontal surface if the coefficient of kinetic friciton between the box

tatuchka [14]

Answer:

Total work done is 2606.08 J.

Explanation:

Given :

Mass of box , m = 23 kg .

Force applied , F = 100 N .

Angle from horizon , $\theta=39^o$ .

Coefficient of kinetic friction , $\mu=0.16$ .

Distance travelled by box , d = 34 m .

Now ,

Total work done = work done by boy + work done by friction.

$W=Fdcos\theta+f_sdcos\theta=Fd-\mu(mg) \\\\W=100\times cos39^o\times 34-0.16\times 23 \times 9.8 \\\\W=77.71\times 34-36.06=2606.08\ J.$

Hence , this is the required solution.

6 0

3 years ago