All categories

2 years ago

Brainliest for correct answer in newtons :)

Physics

2 answers:

Burka [1]2 years ago

6 0

Answer:

120 N

Explanation:

1 N = 1kg

120 N = 120kg

Hope this helps!

Sholpan [36]2 years ago

3 0

Answer:

This is your answer

You might be interested in

PLEASE HELP!! I'll mark brainliest for correct answer.

Tju [1.3M]

1. Car A
2. Car C
3. Car A
4. 1500 J
5. 750 J
6. 750 J
7. Car A was had a height higher relative to the ground compared to all the other cars.
8. mgh
9. B

6 0

2 years ago

A 40-cm-long tube has a 40-cm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. As th

Umnica [9.8K]

Answer:

The frequency of the tuning is 1.065 kHz

Explanation:

Given that,

Length of tube = 40 cm

We need to calculate the difference between each of the lengths

Using formula for length

$\Delta L=L_{2}-L_{1}$

$\Delta L=74.7-58.6$

$\Delta L=16.1\ m$

For an open-open tube,

We need to calculate the fundamental wavelength

Using formula of wavelength

$\lambda=2\Delta L$

Put the value into the formula

$\lambda=2\times16.1$

$\lambda=32.2\ cm$

We need to calculate the frequency of the tuning

Using formula of frequency

$f=\dfrac{v}{\lambda}$

Put the value into the formula

$f=\dfrac{343}{32.2\times10^{-2}}$

$f=1065.2\ Hz$

$f=1.065\ kHz$

Hence, The frequency of the tuning is 1.065 kHz

3 0

2 years ago

Three equal point charges, each with charge 1.45 μCμC , are placed at the vertices of an equilateral triangle whose sides are of

LUCKY_DIMON [66]

Answer:

U = 80.91 J

Explanation:

In order to calculate the electric potential energy between the three charges you use the following formula:

$U=k\frac{q_1q_2}{r_{1,2}}$ (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q1: q2 charge

r1,2: distance between charges 1 and 2.

For the three charges you have:

$U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}$ (2)

You use the fact that q1=q2=q3=q and that the distance between charges are equal. Then, in the equation (2) you have:

q = 1.45μC = 1.45*10^-6C

r = 0.700mm = 0.700*10^-3m

$U_T=3k\frac{q^2}{r}=3(8.98*10^9Nm^2/C^2)\frac{(1.45*10^{-6}C)}{0.700*10^{-3}m}\\\\U_T=80.91J$

The electric potential energy between the three charges is 80.91 J

7 0

2 years ago

A spring stretches 1.68cm vertically when a 2.50kg object is suspended from it.Find the distance (in cm) the spring stretches if

Llana [10]

Answer:

2.96 cm

Explanation:

By Hook's law

Force(F) = Spring constant(k) × Extension(d)

F = k × d

Force is the weight of the object, F = W = mg

So we get, mg = kd ⇒ m ∝ d

2.5 ∝ 1.68 --------------(1)

4.4 ∝ d' --------------(2)

From (1) & (2), 4.4/2.5 = d'/1.68

d' = 2.96 cm ⇒ the required extension.

7 0

3 years ago

Place these waves in order from shortest wavelength to longest.

dsp73

Answer:

Ocean waves, sound waves, and light waves.

Explanation:

5 0

2 years ago

Read 2 more answers