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nignag [31]
2 years ago
5

When a car is 100 meters from its starting position traveling at 60.0 m/s., it starts braking and comes to a stop 350 meters fro

m its starting position in 8.3 s. Calculate the acceleration.
Physics
1 answer:
NISA [10]2 years ago
4 0
Remember your kinematic equations for constant acceleration. One of the equations is x_{f} =  x_{i} +  v_{i}(t) + \frac{1}{2} at^{2}, where x_{f} = final position, x_{i} = initial position, v_{i} = initial velocity, t = time, and a = acceleration. 

Your initial position is where you initially were before you braked. That means x_{i} = 100m. You final position is where you ended up after t seconds passed, so x_{f} = 350m. The time it took you to go from 100m to 350m was t = 8.3s. You initial velocity at the initial position before you braked was v_{i} = 60.0 m/s. Knowing these values, plug them into the equation and solve for a, your acceleration:
350\:m = 100\:m + (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\
250\:m = (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\
250\:m = 498\:m +34.445\:s^{2}(a)\\
-248\:m = 34.445\:s^{2}(a)\\
a \approx -7.2 \: m/s^{2}

Your acceleration is approximately -7.2 \: m/s^{2}.
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The electrical potential energy of a charge q located at a point at potential V is given by
U=qV
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\Delta U = q (V_2 -V_1)=q \Delta V

In our problem, the electron (charge e) must travel across a potential difference V. So the energy it will lose traveling from the metal to the detector will be equal to 
\Delta U = e V
Therefore, if we want the electron to reach the detector, the minimum energy the electron must have is exactly equal to the energy it loses moving from the metal to the detector:
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5 0
2 years ago
A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly abov
Rudik [331]

Answer:

Apparent depth (Da) = 60.15 cm (Approx)

Explanation:

Given:

Distance from fish (D) = 80 cm

Find:

Apparent depth (Da)

Computation:

We know that,

Refractive index of water (n2) = 1.33

So,

Apparent depth (Da) = D(n1/n2)

Apparent depth (Da) = 80 (1/1.33)

Apparent depth (Da) = 60.15 cm (Approx)

5 0
2 years ago
A 3.00-kg model airplane has velocity components of 5.00 m/s due east and 8.00 m/s due north. What is the plane’s kinetic energy
GalinKa [24]

Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

Mass of the model airplane, m = 3 kg

Velocity component, v₁ = 5 m/s (due east)

Velocity component, v₂ = 8 m/s (due north)

Let v is the resultant of velocity. It is given by :

v=\sqrt{v_1^2+v_2^2}

v=\sqrt{5^2+8^2}=9.43\ m/s

Let E is the kinetic energy of the plane. It is given by :

E=\dfrac{1}{2}mv^2

E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2

E = 133.38 Joules

So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.

5 0
2 years ago
Read 2 more answers
A worker stands still on a roof sloped at an angle of 35° above the horizontal. He is prevented from slipping by static friction
aleksley [76]

Answer:

99.63 kg

Explanation:

From the force diagram

N = normal force on the worker from the surface of the roof

f = static frictional force = 560 N

θ = angle of the slope = 35

m = mass of the worker

W = weight of the worker = mg

W Cosθ = Component of the weight of worker perpendicular to the surface of roof

W Sinθ = Component of the weight of worker parallel to the surface of roof

From the force diagram, for the worker not to slip, force equation must be

W Sinθ = f

mg Sinθ = f

m (9.8) Sin35 = 560

m = 99.63 kg

5 0
2 years ago
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topjm [15]

Answer: Weight only.

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If a rock is transported from the moon to the earth, the mass is constant for the object but the weight will depends on the locations of the object. The gravitational acceleration would change because the radius and mass of the Moon is different from the Earth.

Thus, the object (rock) has <em>mass, m</em> both on the surface of the Earth and the surface of the Moon; but it will <em>weight</em> much less on the surface of the Moon as the Moon's surface gravity is 1/6 of the Earth.

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2 years ago
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