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seraphim [82]
2 years ago
7

8x = -6. What does x equal?

Physics
2 answers:
pentagon [3]2 years ago
8 0

Answer:

x=-3/4

Explanation:

allsm [11]2 years ago
4 0

Answer:

x= -3/4

Explanation:

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A car moving at 10.0 m/s encounters a depression in the road that has a circular cross-section with a radius of 30.0 m. What is
Dennis_Churaev [7]

Answer:

F = 789 Newton

Explanation:

Given that,

Speed of the car, v = 10 m/s

Radius of circular path, r = 30 m

Mass of the passenger, m = 60 kg

To find :

The normal force exerted by the seat of the car when the it is at the bottom of the depression.

Solution,

Normal force acting on the car at the bottom of the depression is the sum of centripetal force and its weight.

N=mg+\dfrac{mv^2}{r}

N=m(g+\dfrac{v^2}{r})

N=60\times (9.81+\dfrac{(10)^2}{30})

N = 788.6 Newton

N = 789 Newton

So, the normal force exerted by the seat of the car is 789 Newton.

6 0
2 years ago
A particular conductor has 3.0 × 10^27 mobile electrons per cubic meter. The material is in the shape of a cylinder of length 6.
irina1246 [14]

Answer:

Explanation:

Given

Length of cylinder L=6 cm

diameter of cylinder d=1 cm

Area of cross-section A=\frac{\pi\times d^2}{4}

A=\frac{\pi \times 10^{-4}}{4}=7.855\times 10^{-5} m^2

Emf =2.5 V

current I=5 mA

Resistance=\frac{V}{I}=\frac{2.5}{5\times 10^{-3}}

R=500 \Omega

R is also given by

R=\rho \frac{L}{A}

where \rho =resistivity

\rho =\frac{RA}{L}=\frac{500\times 7.85\times 10^{-3}}{6}

\rho =0.654 \Omega -m        

6 0
2 years ago
Read 2 more answers
Who said an atom is like a ball with no charge?
g100num [7]

Answer:

Thomson

Explanation:

Thomson in 1904 soon after the discovery of the electron, but before the discovery of the atomic nucleus, the model tried to explain two properties of atoms then known: that electrons are negatively-charged particles and that atoms have no net electric charge.

6 0
3 years ago
Please help me i dont get it
Dima020 [189]

The answers are:

<u>39:</u>

a - 1 mile

b - 0 miles

<u>40:</u>

a=-4\frac{m}{s^{2} }

<u>41:</u>

Displacement=3.60 miles\\Distance=5 miles

Why?

Solving 39:

To answer the questions, we need to remember that distance and displacement are different things. Distance refers to the total "ground" covered during motion, while displacement refers to how far is the object/body from its starting point.

So,

<u>39:</u>

a - the total distance traveled corresponds to the length of the track which is 1 mile.

b - The displacement is equal to 0 because they finished at the starting point,  the distance between the starting and the finishing point is equal to 0.

<u>40:</u>

We can solve the problem using one of the given equations:

a=\frac{v_f-v_i}{t}

Since we know all the information, we just need to substitute it into the equation:

a=\frac{v_f-v_i}{t}\\\\a=\frac{2\frac{m}{s} -16\frac{m}{s} }{3.5s}=\frac{-14\frac{m}{s} }{3.5s}\\\\a=-4\frac{m}{s^{2} }

So, the car's acceleration was -4m/s2 (the car was reducing its speed)

<u>41:</u>

We can solve the problem using one of the given equations (Pythagorean Theorem):

c^{2}=a^{2}+b^{2}

Displacement=\sqrt{(3mi(North))^{2}+(2miles(East))^{2}}\\\\Displacement=\sqrt{9+4}=\sqrt{13mi^{2}}=3.60miles

The distance will be:

Distance=3mi+2mi=5miles

Have a nice day!

3 0
2 years ago
to masses 7 kg and 12 KG are connected at the two ends of a light inextensible string that passes over a fictional Pulley using
Alex17521 [72]

Tension in the string when the masses are released ​is 88.42  N

Acceleration of masses is \bold{a=2.578 m/sec^2}

Explanation:  

Given:

Mass ,m1 = 12

Mass , m2 = 7  

g = 9.8m/s^2

To Find :

Tension  in the string=?

Acceleration of masses=?

Solution:

For mass M_1  

M_1 a=T-M_1 g--------------------(1)

For mass M2

M_2 a=T-M_2 g---------------------(2)

Adding equation (1) and (2)

(M_1+M_2)a=(M_2-M_1)g

Finding Acceleration:

Acceleration is given by

a=(M_2-M_1 )/(M_1+M_2) g

Substituting the values,

a=\frac{(12-7)(9.8)}{(7+12)}

a=\frac{(5)(9.8)}{19}

a=\frac {49}{19}

a=2.578 m/sec^2

Finding Tension:

From  Equation 1

M_1 a=T-M_1 g

Tension can be

T=M_1 a+M_1 g

T=(7)(2.578) + 7(9.8)

T=(17.99)+(68.6)

T=86.59 N

8 0
3 years ago
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