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belka [17]
3 years ago
15

Tangential velocity 2. Parabolic pathway 3. Projectile 4. Centripetal acceleration 5. Centripetal force a. acceleration towards

the center caused by the centripetal force b. a force which keeps a body moving with a uniform speed along a circular path and is directed along the radius towards the center c. a curved path followed by projectiles d. an object projected through space, traveling in two dimensions, that accelerates vertically due to gravity e. the instantaneous velocity of a body moving in a circular path
Physics
1 answer:
Stels [109]3 years ago
4 0

1. Tangential velocity:

<em>e) the instantaneous velocity of a body moving in a circular path.</em>

2. Parabolic pathway

<em>c. a curved path followed by projectiles</em>

3. Projectile

<em>d) an object projected through space, traveling in two dimensions, that accelerates vertically due to gravity.</em>

4. Centripetal acceleration

<em>a) acceleration towards the center caused by the centripetal force</em>

5. Centripetal force

<em>b) a force which keeps a body moving with a uniform speed along a circular path and is directed along the radius towards the center</em>

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At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the l
Ksju [112]

Answer:

By convention a negative torque leads to clockwise rotation and a positive torque leads to counterclockwise rotation.

here weight of the child =21kgx9.8m/s2 = 205.8N

the torque exerted by the child Tc = - (1.8)(205.8) = -370.44N-m ,negative sign is inserted because this torque is clockwise and is therefore negative by convention.

torque exerted by adult Ta = 3(151) = 453N , counterclockwise torque.

net torque Tnet = -370.44+453 =82.56N , which is positive means counterclockwise rotation.

b) Ta = 2.5x151 = 377.5N-m

Tnet = -370.44+377.5 = 7.06N-m , positive ,counterclockwise rotation.

c)Ta = 2x151 = 302N-m

Tnet = -370.44+302 = -68.44N-m, negative,clockwise rotation.

5 0
3 years ago
Write the two variables that are dependent on the force of the charge
Murljashka [212]

Answer:

the distance between the particles and also the amount of electric charge they carry.

Explanation:

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8 0
2 years ago
Give examples of plants that can reproduce by vegetative reproduction.​
marishachu [46]

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Some examples of vegetative propagation are farmers creating repeated crops of apples, corn, mangoes or avocados through asexual plant reproduction rather than planting seeds. Vegetative propagation can be accomplished from side-shoots, slips, stems and sections of tubers, bulbs or rhizomes.

Explanation:

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3 years ago
A single-turn current loop, carrying a current of 4.03 A, is in the shape of a right triangle with sides 68.1, 151, and 166 cm.
SashulF [63]

Answer:

Part a)

F = 0

Part b)

F = 0.25 N

Part c)

F = 0.25 N

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

F_{net} = 0

Explanation:

As we know that force on a current carrying wire is given as

\vec F = i(\vec L \times \vec B)

now we have

Part a)

current in side 166 cm and magnetic field is parallel

so we have

F = i(\vec L \times \vec B)

here we know that L and B is parallel to each other so

F = 0

Part b)

For 68.1 cm length wire we have

F = iLB sin\theta

here we know that

cos\theta = \frac{68.1}{166}

\theta = 65.8

so we have

F = (4.03)(0.681)(99.3 \times 10^{-3})sin65.8

F = 0.25 N

Part c)

For 151 cm length wire we have

F = iLB sin\phi

here we know that

cos\phi = \frac{151}{166}

\theta = 24.5

so we have

F = (4.03)(1.51)(99.3 \times 10^{-3})sin24.5

F = 0.25 N

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

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4 0
3 years ago
Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and
strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance R_e defined by the formula:

\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\

and when this is combined with the third resistor in series, the equivalent resistance (R''_e) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}

The problem states that the difference between the equivalent resistances in both circuits is given by:

R''_e=R_e+630 \,\Omega

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega

8 0
3 years ago
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