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2 years ago

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A student is getting ready to go on a bike ride, but her bike back tire is flat! She connects a pump to the valve and adds air t

o the tire until it is fully inflated and ready to ride. After the addition of air to the tire, the air pressure inside the tire is now __________ the air pressure outside the tire. Fill in the blank by choosing one answer below.\

1 answer:

Mariana [72]2 years ago

4 0

Wheres the answer below?

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Two particles with oppositely signed charges nC are placed at two of the vertices of an equilateral triangle with side length 3

babymother [125]

The magnitude of the electric field at the third vertex of the triangle is determined as zero.

<h3>Electric field at the third vertex of the triangle </h3>

The electric field at the third vertex of the equilateral triangle due to the other charges placed on the first and second vertices is calculated as follows;

E = E(13) + E(23)

E = (kq₁)/r² + (kq₂)/r²

where;

q1 is positive charge
q2 is negative charge

E = (kq₁)/r² - (kq₂)/r²

E = 0

Thus, the magnitude of the electric field at the third vertex of the triangle is determined as zero.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

5 0

2 years ago

A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.

pshichka [43]

Answer:

A. 4,9 m/s2

B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package. Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

$\sum F_y=m_1*a_m_i_n = T-m_1*g$

If the package is barely lifted, that means that T=m_2*g; then:

$\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g$

Solving the equation for a_mín, we have:

$a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2$

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: $\sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a$

For the package: $\sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a$

The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package: $\sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a$

We have two unknowns and two equations, so we can proceed. We can match both tensions and have:

$m_1*a+m_1*g=m_2*g-m_2*a$

Solving a, we have

$(m_1+m_2)*a =(m_2 - m1)*g\\\\a=((m_2-m_1)/(m_1+m_2))*g \rightarrow a=((15kg-10kg)/(10kg+15kg))*9,8 m/s^2\\\\a= 2,0 m/s^2$

We can then replace this value of a in one for the sums of force and find the tension T:

$T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N$

5 0

3 years ago

Ivan drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 7 hours. When Ivan drove h

lesya692 [45]

Answer:252 miles

Explanation:

Given

During his way to mountain it took 7 hr to drive

and during his return trip it took 4 hr to return

Let x be the distance between home and mountain

average speed for return is 27 miles per hour faster than his former trip

let v be the speed on his way to mountain thus v+27 is his return speed

thus $7=\frac{x}{v}$ ----1

for return trip

$4=\frac{x}{v+27}$ -----2

divide 1 & 2

$\frac{7}{4}=\frac{x\cdot (v+27)}{v\cdot x}$

$7v=4v+4\cdot 27$

$3v=4\cdot 27$

$v=36 mph$

thus $x=7\times 36=252\ miles$

7 0

3 years ago

A circular coil of wire having a diameter of 20.0 cm and 3000 turns is placed in the earth's magnetic field with the normal of t

Bess [88]

Explanation :

It is given that,

Diameter of the coil, d = 20 cm = 0.2 m

Radius of the coil, r = 0.1 m

Number of turns, N = 3000

Induced EMF, $\epsilon=1.5\ V$

Magnitude of Earth's field, $B=10^{-4}\ T$

We need to find the angular frequency with which it is rotated. The induced emf due to rotation is given by :

$\epsilon=NBA\omega$

$\omega=\dfrac{\epsilon}{NBA}$

$\omega=\dfrac{1.5}{3000\times 10^{-4}\times \pi (0.1)^2}$

$\omega=159.15\ rad/s$

So, the angular frequency with which the loop is rotated is 159.15 rad/s. Hence, this is the required solution.

3 0

3 years ago

Which combination of a wire coil and a core would make the weakest

Ilya [14]

Answer: B

Explanation:

7 0

3 years ago

Read 2 more answers