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3 years ago

If you can’t open the lid of a jelly jar what should you do?

Physics

2 answers:

tangare [24]3 years ago

7 0

Hit in the center of the top, and it will let some air out, that will make it easier.

9966 [12]3 years ago

6 0

If I can't open the lid of a jelly jar, I'd keep trying and if I can't open the lid of a jelly jar after the MANY tries I took, I'd ask for help.

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Astronomers discover a bright X-ray source in the Milky Way. They see no counterpart in the optical bands, but it has another st

VladimirAG [237]

Answer:

a. a black hole

Explanation:

X-ray emission from the central degrees of the Milky Way Bright X-ray emission traces the coherent edge brightened shell-like feature, dubbed the northern chimney, located north of Sgr A* and characterized by a diameter of about 160 pc. On the opposite side, the southern chimney appears as a bright linear feature. Bright X-ray emission is observed at high latitude

8 0

2 years ago

Which of the following wavelengths will produce standing waves on a string that is 3.5 m long?

denpristay [2]

In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:

$\lambda=\frac{2}{n} L$

The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:

$\lambda=\frac{2}{1} \cdot 3.5 m=7.0 m$

The wavelength of the 2nd harmonic is:

$\lambda=\frac{2}{2} \cdot 3.5 m=3.5 m$

The wavelength of the 4th harmonic is:

$\lambda=\frac{2}{4} \cdot 3.5 m=1.75 m$

It is not possible to find any integer n such that $\lambda=5 m$ , therefore the correct options are A, B and D.

3 0

2 years ago

Read 2 more answers

In a generator, as the magnet spins, opposite poles of the magnet push the electrons in opposite directions. This back-and-forth

Natali [406]

C) alternating current .
<span>
B)direct current </span>

6 0

3 years ago

Read 2 more answers

Two point charges 3q and −8q (with q > 0) are at x = 0 and x = L, respectively, and free to move. A third charge is placed so

riadik2000 [5.3K]

Answer:

Explanation:

The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.

So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.

force on it due to rest of the charges will be equal and opposite so

k3q Q / x² =k 8q Q / (L+x)²

8x² = 3 (L+x)²

2√2 x = √3 (L+x)

2√2 x - √3 x = √3 L

x(2√2 - √3 ) = √3 L

x = √3 L / (2√2 - √3 )

Let us consider the balancing force on 3q

force on it due to -Q and -8q will be equal

kQ . 3q / x² = k3q 8q / L²

Q = 8q (x² / L²)

so charge required = - 8q (x² / L²)

and its distance from x on negative x side = √3 L / (2√2 - √3 )

3 0

3 years ago

A race car reaches the finish line and the driver slows down to come to a stop. If it takes the car 10 seconds to stop at a cons

boyakko [2]

Answer:

53.64 m/s

Explanation:

Applying,

a = (v-u)/t............. Equation 1

Where a = acceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.

make u the subject of the equation

u = v-at............. Equation 2

From the question,

Given: a = -12 mph/s = -5.364 m/s², t = 10 seconds, v = 0 m/s (comes to stop)

Substitute these values into equation 2

u = 0-(-5.364×10)

u = 0+53.64

u = 53.64 m/s

6 0

2 years ago