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3 years ago

14

Which layer of the atom sphere contacts the surface of earth

1 answer:

Keith_Richards [23]3 years ago

4 0

The troposphere is the layer of the atmosphere which contacts the surface of the Earth. This layer is <span>0 to 12 km (0 to 7 miles).

Faith xoxo</span>

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What can make a stronger electric or magnetic field?

rusak2 [61]

Answer:

add more coils

Explanation:

this increases the strength of the magnetic field

5 0

2 years ago

2. When a fire truck moves away from you, you hear the pitch of the siren go down. This is

Yanka [14]

the Doppler effect. (I don't know how to explain it lol)

8 0

2 years ago

Read 2 more answers

What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force $F_{1}=18\ N$

Force $F_{2}=26\ N$

Force $F_{3}=14\ N$

We need to calculate the torque due to force F₁

Using formula of torque

$\tau_{1}=-F_{1}d_{1}$

$\tau_{1}=-F_{1}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{1}=-18\times\dfrac{0.2}{2}$

$\tau_{1}=-1.8\ N-m$

We need to calculate the torque due to force F₂

Using formula of torque

$\tau_{2}=F_{2}d_{2}$

$\tau_{2}=F_{2}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{2}=26\times\dfrac{0.2}{2}$

$\tau_{2}=2.6\ N-m$

We need to calculate the torque due to force F₃

Using formula of torque

$\tau_{3}=F_{3}d_{3}$

$\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{3}=0.1(14\sin45+14\cos45)$

$\tau_{3}=1.92\ N-m$

We need to calculate the net torque on the square plate

$\tau=\tau_{1}+\tau_{2}+\tau_{3}$

$\tau=-1.8+2.6+1.92$

$\tau=2.72\ N-m$

Hence, The net torque on the square plate is 2.72 N-m.

3 0

3 years ago

A 1210 kg rollercoaster car is

ratelena [41]

Answer: 4.98 m/s

Explanation:

You solve these kinetic energy, potential energy problems by using the fact P.E.+ K.E. = a constant as long as friction is ignored.

PEi = 0 in this case

KEi = ½mVi² = PEf+KEf = mghf + ½mVf²

½1210*8.31² = 1210*9.8*2.26 + ½1210*Vf²

½1210*Vf² = ½1210*8.31² - 1210*9.8*2.26

Vf² = 8.31² - 2*9.8*2.26 = 4.98² so Vf = 4.98m/s

3 0

2 years ago

Two vehicles, a car and a truck, leave an intersection at the same time. the car heads east at an average speed of 50 miles pe

olya-2409 [2.1K]

The car heads east at an average speed of 50 miles per hour from the intersection point towards East. The truck heads east at an average speed of 60 miles per hour from the intersection point towards South.

The distance of car from the intersection point after t hours is $50t$ .

The distance of truck from the intersection point after t hours is $60t$ .

Since these distances are perpendicular to each other, distance apart d (in miles) at the end of t hours is

$d=\sqrt{(50t)^2+(60t)^2} \\ d=10\sqrt{61} t\\ d=78.1t$

Thus the distance apart is $d=78.1t \;miles$

5 0

3 years ago