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The drawing shows a large cube (mass = 28.6 kg) being accelerated across a horizontal frictionless surface by a horizontal force

MrRissso [65]

Answer:

P= 454.11 N

Explanation:

Since P is the only horizontal force acting on the system, it can be defined as the product of the acceleration by the total mass of the system (both cubes).

$P= (M+m)*a\\a = \frac{P}{28.6 +4.3}\\a = \frac{P}{32.9}$

The friction force between both cubes (F) is defined as the normal force acting on the smaller cube multiplied by the coefficient of static friction. Since both cubes are subject to the same acceleration:

$F = m * a*\mu \\F= 4.3*0.710*\frac{P}{32.9}\\F=3.053*\frac{P}{32.9}$

In order for the small cube to not slide down, the friction force must equal the weight of the small cube:

$3.053*\frac{P}{32.9} = 4.3 * g\\\\P = \frac{4.3*9.8*32.9}{3.053} \\P= 454.11 N$

The smallest magnitude that P can have in order to keep the small cube from sliding downward is 454.11 N

8 0

3 years ago

plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 85.0 nC . The plates are in vacuu

tensa zangetsu [6.8K]

Answer:

12500 V

Explanation:

The electric field in the gap of a parallel-plate capacitor is uniform, so the following relationship between electric field strength, potential difference and distance can be used:

$\Delta V = E d$

where

$\Delta V$ is the potential difference between the plates

E is the electric field strength

d is the distance between the plates

For the capacitor in this problem, we have

$E=5.00\cdot 10^6 V/m$

$d = 2.50 mm = 2.50\cdot 10^{-3} m$

Substituting, we find

$\Delta V = (5.00\cdot 10^6)(2.50\cdot 10^{-3})=12500 V$

3 0

3 years ago

A 15-turn circular wire loop with a radius of 3.0 cm is initially in a uniform magnetic field with a strength of 0.5 T. The fiel

lana66690 [7]

To solve this problem it is necessary to apply the definition given in Faraday's law in a solenoid for which it is noted that

$\epsilon = - d\frac{\phi_B}{dt}$

$\epsilon = -NA\frac{dB}{dt}$

Where,

N = Number of loops

A = Cross sectional Area

B = Magnetic Field

$\epsilon = (15)(\pi(0.03)^2)\frac{0-0.5}{0.1}$

$\epsilon = 0.212V$

$\epsilon = 0.21V$

Therefore the correct answer is A.

6 0

3 years ago

If we assume that, in the African savanna, the predator population remains constant, what other, nonmobile, factor would influen

musickatia [10]

On usa test prep the answer is c

7 0

3 years ago

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Which statement supports Newton’s first law of motion?

Ipatiy [6.2K]

Answer: The statement that supports Newton's first law of motion is the one that says that planets can move at a varying speed due to forces exerted in space

6 0

3 years ago

Read 2 more answers