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3 years ago

13

What type of clothing is most suitable to facilitate this? A. a pair of pants with a cord B. a flame-resistant pair of pants C.

a pair of pants with an elastic waistband D. a pair of pants with buttons

1 answer:

Tasya [4]3 years ago

8 0

Answer:

A

Explanation:

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An electron in the beam of a cathod-ray tube is accelerated by a potential difference of 2.12 kV . Then it passes through a regi

son4ous [18]

Answer:

B=9.1397*10^-4 Tesla

Explanation:

To find the velocity first we put kinetic energy og electron is equal to potential energy of electron

K.E=P.E

$\frac{1}{2}*m*v^{2} =e*V$

where :

m is the mass of electron

v is the velocity

V is the potential difference

$v=\sqrt{\frac{2*e*V}{m} }$ eq 1

Radius of electron moving in magnetic field is given by:

$R=\frac{m*v}{q*B}$ eq 2

where:

m is the mass of electron

v is the velocity

q=e=charge of electron

B is the magnitude of magnetic field

Put v from eq 1 into eq 2

$R=\frac{m*\sqrt{\frac{2*e*V}{m} } }{e B}$

$B=\sqrt{\frac{2*m*V}{e*R^{2} } }$

$B=\sqrt{\frac{2*(9.31*10^{-31})*(2.12*10^{3}) }{(1.60*10^{-19})*(0.170)^{2} } }$

B=9.1397*10^-4 Tesla

3 0

3 years ago

Which vector shows the direction of the centripetal acceleration at this point

Mademuasel [1]

Answer:

It's B

Explanation:

6 0

3 years ago

You can enter compound units that are combinations of other units that are multiplied together.Torque can be calculated by multi

Paraphin [41]

Answer:

3.0 x 10¹ Nm

Explanation:

Torque = F x r

Where F is force applied and r is perpendicular distance from pivot point . r

is also called lever arm

Here F = 15 N and r = 2.0 m

Torque

= 15 N X 2.0 m

= 3.0 10¹ Nm.

4 0

3 years ago

A man jogs at a speed of 1.6 m/s. His dog

FromTheMoon [43]

I believe it is
1.6x=2.7(x-1.8)
1.1x=2.7*1.8
x~4.4
4.4*1.6
~7.1m

5 0

3 years ago

Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin

likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0 and T - mgcos57 - F = m(0) = 0

Mg - T = 0 (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

6 0

3 years ago