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3 years ago

6

"Force" can be defined as a push or a O pull O grab

2 answers:

valkas [14]3 years ago

7 0

Answer:

pull is your answer.

I HOPE your day goes great.

Nady [450]3 years ago

4 0

Answer:

pull

is your answer please give me some thanks

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Please someone help me with this!!!

exis [7]

Answer: Gravitational potential energy changes.

Explanation: This is because depending on the amount of mass in an object that’s the amount of gravity pulling you down to the center of the earth

5 0

2 years ago

Help me on this question

nikitadnepr [17]

Adjust the height of the wooden rod so that it just touches the surface of the water. Switch on the lamp and motor and adjust the speed of the motor until low frequency waves can be clearly observed... Count the number of waves passing a point in ten seconds then Divide by ten to record frequency.

6 0

3 years ago

When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0

lora16 [44]

Answer:

$d=0.165m$

Explanation:

Given

$m=0.25kg$ , $x_{1}=5cm*\frac{1m}{100cm}=0.05m$ , $x_{2}=4cm*\frac{1m}{100cm}=0.04m$ , $v=2\frac{rev}{s}$

The tension of the spring is

$F_{k}=K*x_{1}=m*g$

$K=\frac{m*g}{x_{1}}$

$K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m$

The force in the spring is equal to centripetal force so

$F_{c}=\frac{m*v^2}{r}$

$v=w*r=2\pi*r$

But Fc is also

Fc=KxΔr

$F_{c}=K*(r-x_{2})$

Replacing

$m*4\pi^2*r=K*(r-x_{2})$

$0.25kg*4\pi^2*r=49*(r-0.04m)$

$r=0.205m$

total distance is

$d=0.205-0.04=0.165m$

3 0

3 years ago

If a leaf falls from a tree, has work been done on the leaf? Explain.

mihalych1998 [28]

Answer:

Hope this helps!

5 0

3 years ago

What would be the answer for this and how?

Veronika [31]

Answer:

B. 6 cm

Explanation:

First, we calculate the spring constant of a single spring:

$k = \frac{F}{\Delta x}\\$

where,

k = spring constant of single spring = ?

F = Force Applied = 10 N

Δx = extension = 4 cm = 0.04 m

Therefore,

$k = \frac{10\ N}{0.04\ m}\\k = 250\ N/m\\$

Now, the equivalent resistance of two springs connected in parallel, as shown in the diagram, will be:

$k_{eq} = k + k\\k_{eq} = 2k = 2(250\ N/m)\\k_{eq} = 500\ N/m\\$

For a load of 30 N, applying Hooke's Law:

$\Delta x = \frac{F}{k_{eq}}\\\\\Delta x = \frac{30\ N}{500\ N/m}\\\\\Delta x = 0.06\ m = 6\ cm\\$

Hence, the correct option is:

<u>B. 6 cm</u>

7 0

3 years ago