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3 years ago

13

Each of the boxes starts at rest and is then pulled for 2.0 m across a level, frictionless floor by a rope with the noted force.

Which box has the highest final speed

1 answer:

Yuki888 [10]3 years ago

3 0

Answer:

Explanation:

d

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2.85 A police car is traveling at a velocity of 18.0 m/s due north, when a car zooms by at a constant velocity of 42.0 m/s due n

katrin2010 [14]

Answer:

11.1 s

Explanation:

Speed of the police car as given = v = 18 m/s

Speed of the car = V = 42 m/s

Reaction time = t = 0.8 s

Distance traveled by the police car during the reaction time = d₁= 0.8 x 18 = 14.4 m

Distance traveled by speeding car = d₂ =0.8 x 42 = 33.6 m

Acceleration of the police car = a = 5 m/s/s

The police car can catch the speeding car only if it travels a distance equal to the speeding car in a time t.

Distance traveled by the police car = D = d₁ + v t +0.5 at², according to the kinematic equation.

⇒ D = 14.4 + 18 t + 0.5 (5) t²

⇒ D = 14.4 + 18 t+2.5 t² → (1)

For the speeding car, distance traveled is D = 33.6 + 42 t, since it is constant velocity. Substitute for D from the above equation (1).

⇒ 14.4 + 18 t+2.5 t²= 33.6 + 42 t

⇒ 2.5 t² -24 t - 19.2 = 0

⇒ t = 10.3 s

Total time = t +0.8 s

⇒ Time taken for the police car to reach the speeding car = 10.3+0.8= 11.1 s

5 0

2 years ago

Suppose a 4,000-kg elephant is hoisted 20 m above Earth’s surface. Use a calculator and follow the steps below to find the eleph

stiv31 [10]

GPE = 78,380 J
w = 39,240 N

First list what you know. You know the elephants mass and it’s height. You also know gravity on Earth. I will use g = 9.81.
m = 4,000 kg
h = 20 m
g = 9.81 m/s^2

You need to find the elephants weight. Weight = mass x gravity
w = mg
w = (4000 kg)(9.81 m/s^2)
w = 39,240 N (N = newtons)

Now, knowing the elephants weight, you can calculate its GPE.
Gravitational Potential Energy = weight x height

GPE = wh
GPE = (39,240N)(20m)
GPE = 78,380 J (J = joules)

4 0

3 years ago

How much work is done against gravity when lowering a 16 kg box 0.50 m? (g = 9.8 m/s2)

leonid [27]

Answer:

The work done against gravity is 78.4 J

Explanation:

The work is calculated by multiplying the force by the distance that the

object moves

W = F × d, where W is the work , F is the force and d is the distance

The SI unit of work is the joule (J)

We need to find the work done against gravity when lowering a

16 kg box 0.50 m

→ F = mg

→ m = 16 kg, and g = 9.8 m/s²

Substitute these value in the rule

→ F = 16 × 9.8 = 156.8 N

→ W = F × d

→ F = 156.8 N and d = 0.50

Substitute these values in the rule

→ W = 78.4 J

<em>The work done against gravity is 78.4 J</em>

6 0

3 years ago

In a mail-sorting facility, a 2.50-kg package slides down an inclined plane that makes an angle of 20.0° with the horizontal. Th

lawyer [7]

Answer:

The coefficient of kinetic friction is 0.382.

Explanation:

Given:

Angle of inclination is, $\theta=20.0$ °

Mass of package is, $m=2.50\ kg$

Initial speed of package is, $u=2.00\ m/s$

Final speed of the package at the bottom is, $v=0\ m/s$

Distance of travel along the incline is, $d=12.0\ m$

Acceleration due to gravity is, $g=9.8\ m/s^2$

Let the coefficient of kinetic friction be $\mu$ .

Now, the frictional force will be acting along the incline but in the direction opposite to the direction of motion.

So, the net acceleration acting on the package will be up the incline and is equal to:

$a=\mu g\cos\theta-g\sin\theta$ ----------------- 1

Now, using equation of motion, we have:

$v^2-u^2=2ad\\\\0-(2.00)^2=2a(12.0)$

Solving for 'a', we get:

$-4.00=24.0a\\\\a=-\frac{4}{24}=-\frac{1}{6}\ m/s^2$

Now, plug in the value of 'a' in equation (1). This gives,

$\mu g\cos\theta-g\sin\theta=\frac{1}{6}$ ( Neglecting negative sign)

Plug in all the given values and solve for $\mu$ . This gives,

$9.8(-sin(20)+\mu cos(20))=\frac{1}{6}\\\\-0.342+\mu\times 0.94=0.017\\\\0.94\mu=0.342+0.017\\\\0.94\mu=0.359\\\\\mu=\frac{0.359}{0.94}=0.382$

Therefore, the coefficient of kinetic friction is 0.382.

5 0

3 years ago

Calculate the energy released by the electron-capture decay of 5727Co. Consider only the energy of the nuclei (ignore the energy

erma4kov [3.2K]

Answer:

Explanation:

⁵⁷Co₂₇ + e⁻¹ = ²⁷Fe₂₆

mass defect = 56.936296 + .00055 - 56.935399

= .001447 u

equivalent energy

= 931.5 x .001447 MeV

= 1.3479 MeV .

= 1.35 MeV

energy of gamma ray photons = .14 + .017

= .157 MeV .

Rest of the energy goes to neutrino .

energy going to neutrino .

= 1.35 - .157

= 1.193 MeV.

5 0

2 years ago