Which of the following is an example of a force?

In a double-slit experiment, light and dark regions are observed on a screen. what causes a dark region to be observed between t

Colt1911 [192]

According to a source, fringes is the answer. These fringes are what causes dark regions in the double -slit experiment conducted that can be observed in the screen.
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7 0

3 years ago

I throw a football in the x direction with a fore of 124N. The football is 4 kilograms. The football accelerated for 5 seconds.

castortr0y [4]

31 (unit)

acceleration formula

net force (124)

Acceleration = -------------

mass (4)

124÷4=31

6 0

3 years ago

A characteristic in an experiment, such as how much food, how much space, how hot or how cold, is termed a ____________. (spelli

KATRIN_1 [288]

I think the term is a variable?

6 0

3 years ago

Read 2 more answers

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

3 years ago

a crane lifts four pallets of bricks each of which weigh 5000 N. the crane lifts each pallet a height of 30m. the crane takes 4

love history [14]

Answer:

625 W

Explanation:

Applying

P = W/t.................... Equation 1

Where p = power, W = Work, t = time

But,

W = Force (F) × distance (d)

W = Fd........................ Equation 2

Substitute equation 2 into equation 1

P = Fd/t.................... Equation 3

From the question,

Given: F = 5000 N, d = 30 m, t = 4 munites = (4×60) seconds = 240 seconds

Substitute these values into equation 3

P = (5000×30)/240

P = 625 Watt

3 0

3 years ago