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Arada [10]
2 years ago
9

If an experiment involves a large volume of liquid a _______ would most likely be used to hold it.

Physics
2 answers:
Marianna [84]2 years ago
7 0

Answer:

D: Beaker

Explanation:

The beaker has the largest capacity for volume among these options. A beaker is usually used to hold stock liquid, while it is poured into smaller containers, such as test tubes.

Test tubes hold the final amount usually used in the chemical reactions.

Graduated cylinders are usually used to measure certain liquids (mostly from beakers) to obtain precise amounts of a substance.

Hot plates are used to heat substances and cannot hold liquids.

Grace [21]2 years ago
3 0
The answer is d a beacker
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An electron moving with a velocity v⃗ = 5.0 × 107 m/s i^ enters a region of space where perpendicular electric and a magnetic fi
stiv31 [10]

Answer:

Magnetic field, B=2\times 10^{-4}\ T

Explanation:

Given that,

Velocity of electron, v=5\times 10^7\ m/s

It enters  a region of space where perpendicular electric and a magnetic fields are present.

Magnitude of electric field, E=10^4\ V/m

We need to find the magnetic field will allow the electron to go through the region without being deflected.

Magnetic force on the electron, F_m=qvB\ sin\theta.......(1)

Electric force on the electron, F = q E........(2)

From equation (1) and (2) we get:

qvB\ sin\theta=qE

B=\dfrac{E}{v}

B=\dfrac{10^4\ V/m}{5\times 10^7\ m/s}

B = 0.0002 T

or

B=2\times 10^{-4}\ T

Hence, this is the required solution.

3 0
3 years ago
What can be absorbed or released as the result of a chemical reaction?
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3 years ago
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Which process is a form of mechanical weathering?
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2 years ago
An archer shoots an arrow with a velocity of 45.0m/s at an angle of 50.0degrees with the horizontal.An assistant standing on the
garri49 [273]

Answer:

a) u = 30.29 m/s

b) t = 2.09 s

Explanation:

given,

velocity = 45 m/s

angle (θ) = 50°

horizontal velocity = 45 cos 50°

time taken to reach 150 m.

times = \dfrac{150}{45 cos 50^0}

t  =  5.19 s

a) height of arrow

s = u t +\dfrac{1}{2}gt^2

s = v sin \theta \times t+\dfrac{1}{2}gt^2

s = 45 sin 50^0 \times 5.19 -\dfrac{1}{2}\times 9.81\times 5.19^2

s = 46.78 m

v² - u² = 2 g s

u² = 2 × 9.81 × 46.78

u = 30.29 m/s

b) time taken by the apple = \dfrac{u}{g}=\dfrac{30.29}{9.81}

                                             = 3.09 s

time after which it has to be thrown = 5.19-3.09 = 2.1 s

5 0
2 years ago
Can someone help pleaseeee
Eduardwww [97]

Answer:

Motion with constant velocity of magnitude 1 m/s (uniform motion) for 4 seconds in a positive direction and then for 2 seconds uniform motion with constant velocity of magnitude 3 m/s in reverse direction .

Explanation:

The graph shows a constant velocity of 1 m/s for 4 seconds in the positive direction. After that, between 4 seconds and 6 seconds, the object reverses its motion with constant velocity of magnitude 3m/s.

4 0
3 years ago
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