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Ann [662]
3 years ago
12

At time t1 = 14 s, a car is located at 99, 80, 27 m and has velocity 4, 0, −3 m/s. At time t2 = 18 s, what is the position of th

e car? (The velocity is constant in magnitude and direction during this time interval.)
Physics
1 answer:
Korvikt [17]3 years ago
8 0

Answer:

115, 80, 15m

Explanation

t1 = 14s

t2 = 18s

change in time = 4s (18-14)

r(final) = r(initial) + (average velocity) x (change in time)

multiply the average velocity with the change in time

= (4, 0, -3) x 4 = 16, 0, -12

now we'll add this value to the initial position of the car

(99, 80, 27)m + (16, 0, -12)m = (115, 80, 15)m

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An 7.5 × binocular has 3.7-cm-focal-length eyepieces. What is the focal length of the objective lenses? Express your answer to t
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To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,

\mu = \frac{f_0}{f_e}

Here,

\mu = Magnification

f_e = Focal length eyepieces

f_0 = Focal length of the Objective

Rearranging to find the focal length of the objective

f_0 = \mu f_e

Replacing with our values

f_0 = 7.5* 3.7cm

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Therefore the focal length of th eobjective lenses is 27.75cm

5 0
3 years ago
Help pls i need this right now
pantera1 [17]

Answer:

The x-component of F_{3} is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O (1)

Where:

\vec F_{1}, \vec F_{2}, \vec F_{3} - External forces exerted on the ring, measured in newtons.

\vec O - Vector zero, measured in newtons.

If we know that \vec F_{1} = (70.711,70.711)\,[N], \vec F_{2} = (-126.859, 46.173)\,[N], F_{3} = (F_{3,x},F_{3,y}) and \vec O = (0,0)\,[N], then we construct the following system of linear equations:

\Sigma F_{x} = 70.711\,N - 126.859\,N +F_{3,x} = 0\,N (2)

\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N (3)

The solution of this system is:

F_{3,x} = 56.148\,N, F_{3,y} = -116.884\,N

The x-component of F_{3} is 56.148 newtons.

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3 years ago
A skater moves with 15 m/s velocity in a circle of radius circle of radius 30 m. The ice exerts a center force of 450 N. What is
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2 years ago
A 5 kg ball of clay is moving with a speed of 25 m/s directly toward a 10 kg ball of clay which is at rest. The two clay balls c
Dafna11 [192]

Answer:

8.3m/s

Explanation:

Given parameters:

mass of clay ball  = 5kg

Speed of clay ball  = 25m/s

mass of clay ball at rest  = 10kg

speed of clay ball at rest  = 0m/s

Unknown:

Velocity after collision  = ?

Solution:

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           v  = 8.3m/s

7 0
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Which phrase describes a planet’s period of rotation? A. the shape of its orbit B. its tilt relative to the sun C. one complete
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C. one complete spin on its axis because the rotation is referring to the planet's period of rotation. D is called a revolution. B determines the seasons on the planets. A is called an ellipse.
8 0
3 years ago
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