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Tomtit [17]
3 years ago
7

Which graph represents a nonlinear relationship? 40 40 40 40 30 30 30 30 20 از مسیر 20 20 20 10 10 10 10 0 0 0 0 2 4 6 0 2 4 6 0

2. 4. 6 0 2 4 6 0​
Physics
1 answer:
Doss [256]3 years ago
7 0

Explanation:

itu Adalah ekseperiasi Yg sagat

mudah kita megola pelaharan dari mkt

kita igin melamar kerja Apa bila

kita visa mejawab nya degan benar

maka kina di terima jadi kariawan nya

kita terhadap bos harus patuh Dan

tidal pernah melakukan half Yg sgat Kevin

karen tidal boleh

itu lah Yg sya visa di sampaiksn pada kmu brenly

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Paco was driving his scooter west with an initial velocity of 4 m/s. he accelerates at 0.5 m/s2 for 30 seconds. what is his fina
sleet_krkn [62]

The final velocity will be 19 m/sec.Option C is correct.

<h3>What is velocity?</h3>

The change of displacement with respect to time is defined as the velocity.  

Velocity is a vector quantity. it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.

From Newton's first equation of motion;

\rm v = u+at \\\\ v= 4+ 0.5 \times 30 \\\\ v= 19 \ m/sec

The final velocity will be 19 m/sec

Hence, option C is correct.

To learn more about the velocity refer to the link ;

brainly.com/question/862972

#SPJ1

4 0
2 years ago
Two identical soccer balls are rolled toward each other. What will be true after they collide head-on
Juliette [100K]
no because there is no chance that it will possibly hit eachother
6 0
2 years ago
A small block with mass 0.0350kg slides in a vertical circle of radius 0.525m on the inside of a circular track. During one of t
kirill115 [55]

Answer:

w = -0.475N

Explanation:

K.E_{a} + P.E_{a} + W_{fr} = K.E_{b} + P.E_{b}\\K.E = 0.5mv^{2} \\Normal force at point B, N_{B} = 0.665N\\Normal force at point A, N_{A} = 3.85N\\

To get Va and Vb

F = mv_{A} ^{2} /R................(1)\\F = N_{A} - mg.........................(2)\\mv_{A} ^{2} /R =  N_{A} - mg\\v_{A} ^{2} = R (N_{A}/m - g)\\v_{A} = \sqrt{ R (N_{A}/m - g)}

R = 0.525 m

m = 0.0350 kg

g = 9.8 m/s²

v_{A}= \sqrt{0.525(3.85 /0.0350 - 9.8)} \\v_{A} = 7.25 m/s

K.Ea = 0.5 * 0.035 * 7.25²

K.Ea = 0.92 J

Since point A is at the bottom of the path, h = 0 m

P.Ea = 0 m

For Vb

F = mv_{B} ^{2} /R................(1)\\F = N_{B} - mg.........................(2)\\mv_{B} ^{2} /R =  N_{B} - mg\\v_{B} ^{2} = R (N_{B}/m - g)\\v_{B} = \sqrt{ R (N_{B}/m - g)}

N_{B} = 0.665N

v_{B}= \sqrt{0.525(0.665 /0.0350 - 9.8)} \\v_{B} = 2.198 m/s

K.E_{B} = 0.5* 0.035 * 2.198^{2} \\K.E_{B} = 0.085 J

P.E_{B} = mgh_{B} \\h_{B} = A diameter = 2R = 2 * 0.525\\h_{B} = 1.05 m\\P.E_{B} = 0.035 * 9.8 * 1.05\\P.E_{B} =0.36 J

from K.E_{a} + P.E_{a} + W_{fr} = K.E_{b} + P.E_{b}

0.92 + w_{fr} + 0 = 0.085 + 0.36\\ w_{fr} = -0.475J

7 0
4 years ago
(10%) Problem 10: A 7.25-kg bowling ball moving at 9.85 m/s collides with a 0.875-kg bowling pin, which is scattered at an angle
siniylev [52]

Answer

given,

mass of bowling ball = 7.25 Kg

moving speed of the bowling ball = 9.85 m/s

mass of bowling in = 0.875 Kg

scattered at an angle = θ = 21.5°

speed after the collision = 10.5 m/s

angle of the bowling ball

tan \theta_1 = \dfrac{-[m_2v_2Sin \theta_2]}{m_1v_1 - (m_2v_2cos \theta_2)}

tan \theta_1 = \dfrac{-[0.875\times 10.5 \times Sin 21.5^0]}{7.25\times 9.85 - (0.875\times 10.5 \times cos 21.5^0)}

tan \theta_1 = \dfrac{-[3.3672]}{62.86}

tan \theta_1 = 0.0536

\theta_1 =-3.066^0

b) magnitude of final velocity

v = \dfrac{-m_2v_2sin\theta_2}{m_1 sin\theta_1}

v = \dfrac{-0.875 \times 10.5 sin21.5^0}{7.25 sin(-3.066^0)}

v = 8.68 m/s

5 0
3 years ago
The sound of frequency greater than 20,000 Hz is called ....
lapo4ka [179]

Answer:

ultra sound... follow me

3 0
4 years ago
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