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AleksandrR [38]

3 years ago

13

Help yall 13 points!!

1 answer:

Naya [18.7K]3 years ago

3 0

Answer:

Explanation:

12.)

A. Opposite poles attract

B. Same poles repel

13.)

IDK

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the ocean floor is, on average, 4267 m below sea level. What is the pressure in the atmosphere at this depth?

hichkok12 [17]

The pressure at the depth h in the ocean is given by (Stevin's law)
$p= p_0 + \rho g h$
where
$p_0 = 1.0 \cdot 10^5 Pa$ is the atmospheric pressure
and $\rho g h$ is the pressure exerted by the column of water of height h=4267 m, with $\rho = 1000 kg/m^3$ being the water density and $g=9.81 m/s^2$ .
Substituting, we find
$p=1.0 \cdot 10^5 Pa + (1000 kg/m^3)(9.81 m/s^2)(4267 m)=4.20 \cdot 10^7 Pa$
We want to convert this into atmospheres: we know that 1 atm corresponds to the atmospheric pressure at sea level, so $1 atm=1.0\cdot 10^5 Pa$ , therefore we just need to divide by this number:
$p= \frac{4.20 \cdot 10^7 Pa}{1.0 \cdot 10^5 Pa/atm} =420 atm$

7 0

3 years ago

What is the average speed in mph for a car that travels for 5 hours and 20 minutes?

Ymorist [56]

Answer: If you meant 5 miles in 20 minutes than it’s 1 mile in 5 minutes

Explanation:

8 0

3 years ago

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Steam is leaving a 4-L pressure cooker whose operating pressure is 150 kPa. It is observed that the amount of liquid in the cook

Anna71 [15]

Answer:

The mass flow rate is 2.37*10^-4kg/s

The exit velocity is 34.3m/s

The total flow of energy is 0.583 KJ/KgThe rate at which energy leave the cooker is 0.638KW

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7 0

2 years ago

The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

$\sin\theta=\dfrac{3}{5}$

The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

$T=658.33\ N$

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0

3 years ago

1. A 3.1 kg cart is traveling at 7.12 m/s to the right and it has a head on elastic collision with a 11.7 kg cart traveling at 1

Rashid [163]

Answer:

1.03 m/s

Explanation:

I'm too lazy to write the explanation down but my teacher graded this and it was right

6 0

3 years ago