A force of 6.0 N gives a 2.0 kg block an acceleration of 3.0

which symbol in the first law of thermpdynamics represents energy that moves from a hot object to a cooler object?

Marysya12 [62]

The triangle <span>in the first law of thermodynamics, represents energy that moves from a hot object to a cooler object.</span>

4 0

3 years ago

Under which condition would time periods of daylight and darkness be equal everywhere on Earth all year? A. if Earth revolved ar

FromTheMoon [43]

The answer is b because that is the correct one

8 0

3 years ago

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4. Johnny exerts a 3.55 N rightward force on a 0.200-kg box to accelerate it across a low-friction track. If the total resistanc

Anon25 [30]

a) $15.2 m/s^2$

b) 1.96 N

c) 1.96 N

Explanation:

a)

To find the box's acceleration, we have to find first the net force acting on the box in the horizontal direction.

We have:

- The forward force of 3.55 N

- The backward, resistive force of 0.52 N

So, the net force forward is

$\sum F=3.55-0.52=3.03 N$

Now we can find the acceleration by using Newton's second law of motion, which states that:

$\sum F=ma$

where

m = 0.200 kg is the mass of the box

a is its acceleration

And solving for a, we find the acceleration:

$a=\frac{\sum F}{m}=\frac{3.03}{0.200}=15.2 m/s^2$

b)

The gravitational force on an object is the force with which the object is pulled towards the ground by the Earth.

It is given by

$W=mg$

where

m is the mass of the object

g is the gravitational field strength

In this problem we have

m = 0.200 kg is the mass of the box

$g=9.8 m/s^2$ is the gravitational field strength

So, the gravitational force on the box is

$W=(0.200)(9.8)=1.96 N$

c)

The normal force is the reaction force exerted by the floor on the box, in the upward direction.

In order to find the magnitude of this force, we apply Newton's second law of motion along the vertical direction.

We have two forces in this direction:

- The gravitational force, W, downward

- The normal force, N, upward

So the net force is

$\sum F=N-W$

According to Newton's second law,

$\sum F=ma$

However, the box is at rest in the vertical direction, so the vertical acceleration is zero:

$a=0$

This means that the net force is zero:

$\sum F=0$

And so, we can find the normal force:

$N-W=0\\N=W=1.96 N$

4 0

3 years ago

Water emerges straight down from a faucet with a 2.51-cm diameter at a speed of 3.04 m/s. (because of the construction of the fa

Mkey [24]

This is a question on conservation of energy. That is,
mgh + KE1 = KE2
mgh +1/2mv1^2 = 1/2mv2^2
gh + 1/2v1^2 = 1/2v2^2

Where, h = 0.2 m, v1 =3.04 m/s
Therefore,
v2 = Sqrt [2(gh+1/2v1^2)] = Sqrt [2(9.81*0.2 + 1/2*3.04^2)] = 7.26 m/s

Now, Volumetric flow rate, V/time, t = Surface area, A*velocity, v
Where,
V = Av = πD^2/4*3.04 = π*(2.51/100)^2*1/4*3.04 = 1.504*10^-3 m^3/s

At 0.2 m below,
V = 1.504*10^-3 m^3/s = A*7.26
A = (1.504*10^-3)/7.26 = 2.072*10^-4 m^2

But, A = πr^2
Then,
r = Sqrt (A/π) = Sqrt (2.072*10^-4/π) = 0.121*10^-3 m
Diameter = 2r = 0.0162 m = 1.62 cm

3 0

3 years ago

A motor-cycle rider going 90km/hr around a curve with a radius of 100m must lean at an angle to the vertical. The angle at which

BartSMP [9]

Answer:

The angle of inclination of the rider to the vertical is 32.5⁰.

Explanation:

Given;

linear speed of the motor cycle, v = 90 km/hr = 25 m/s

radius of the circle, r = 100 m

The angle of inclination of the rider to the vertical is calculated using banking angle formula;

$\theta = tan^{-1}(\frac{v^2}{rg} )\\\\\theta = tan^{-1} (\frac{25^2}{100 \times 9.8} )\\\\\theta = 32.5 ^0$

Therefore, the angle of inclination of the rider to the vertical is 32.5⁰

3 0

3 years ago