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3 years ago

Please help me very urgent :((

Physics

1 answer:

Stells [14]3 years ago

7 0

Answer:

-150 N

Explanation:

(Newton's second law) F=ma

Sum of forces in Y direction= (+200 N)+(-200 N)= 0...

forces cancel, object does not accelerate up/down

Sum of forces in X direction= (+65 N)+(-65 N)+(-150 N)

= -150 N

notice that the +/- 65 components cancel, leaving a net force of 150 N in the LEFTwards direction (which is typically defined as negative)

Overall, the net force is -150 N

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2001240Determine the specific kinetic energy of a mass whose velocity is 40 m/s, in kJ/kg.

guapka [62]

Answer:

The specific kinetic energy of a mass is 0.8 kJ/kg

Explanation:

Given that,

Velocity = 40 m/s

Specific kinetic energy is the kinetic energy per unit mass.

We need to calculate the specific kinetic energy

Using formula of specific kinetic energy

$K.E=\dfrac{\dfrac{1}{2}mv^2}{m}$

$K.E=\dfrac{v^2}{2}$

Put the value into the formula

$K.E=\dfrac{40^2}{2}$

$K.E=800\ J/kg$

We know that,

1 kJ = 1000 J

or, 1J=0.001 KJ

The specific energy is

$K.E=800\times0.001$

$K.E=0.8\ kJ/kg$

Hence, The specific kinetic energy of a mass is 0.8 kJ/kg

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3 years ago

A construction worker is carrying a load of 40 kg over his head and is walking at a constant velocity if he travels a distance o

Anastasy [175]

Answer:

W = 0

Explanation:

We are given with, a construction worker is carrying a load of 40 kg over his head and is walking at a constant velocity. He travels a distance of 50 m.

The work done by an object is given by :

$W=Fd$

F = ma

So,

$W=mad$

m is mass

a is acceleration

d is displacement

The worker is moving with constant velocity, its acceleration will be 0. So, the work done by the worker is 0.

8 0

4 years ago

A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t

lutik1710 [3]

a) E = 0

b) $3.38\cdot 10^6 N/C$

Explanation:

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

where

E is the electric field

q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

$r_1 = 10 cm\\r_2 = 15 cm$

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$