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astraxan [27]
3 years ago
11

Can someone please help me:(

Physics
1 answer:
finlep [7]3 years ago
8 0

Answer:

<em><u>Incomple</u></em><em><u>te</u></em><em><u> </u></em><em><u>ques</u></em><em><u>tion</u></em><em><u>!</u></em>

you have to provide mass to get acceleration

as Newton's 2nd law, force=mass×accln

now,here force toward D is 100N,if it hav mass m

then accln will be 100/m

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PLEASE ANSWER! WILL MARK BRAINLIEST!
Angelina_Jolie [31]

The first one would be thermal energy

8 0
3 years ago
7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

3 0
3 years ago
A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a
makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]              (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°

The distance r is:

r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m

You replace the values of all parameters in the equation (1):

\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0
3 years ago
Which number is not found on the periodic table​
miv72 [106K]

The letter “j” is never found on the periodic table. As for numbers, there’s an infinite amount

4 0
3 years ago
A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling
Natalka [10]

Answer:

H = 5 m

Explanation:

As the person leaves the slide horizontally so the time taken by the person to hit the water is given as

t = 0.672 s

so we can find the vertical velocity by which person will hit the water using kinematics

v_y = u_y + at

v_y = 0 + (9.81)(0.672)

v_y = 6.6 m/s

now the speed of the person at the end of the slide is given as

v = \frac{L}{t}

v = \frac{5}{0.672}

v_x = 7.44 m/s

now by energy conservation we can find the initial height

mgH = \frac{1}{2}m(v_x^2 + v_y^2)

H = \frac{1}{2g}(v_x^2 + v_y^2)

H = \frac{1}{2(9.81)}(7.44^2 + 6.6^2)

H = 5 m

6 0
3 years ago
Read 2 more answers
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