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3 years ago

The graphs show the motion of an object in both the x and y directions. Classify the motion of this object.

Physics

2 answers:

son4ous [18]3 years ago

8 0

Answer:

Explanation:

Natali5045456 [20]3 years ago

6 0

B It is moving in one dimension

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The liquid contained in batteries that conducts electric charge is called the: solution, electrolyte, suspension, colloid

blondinia [14]

Answer:

The liquid contained in batteries that conducts electric charge is called the. electrolyte.

7 0

2 years ago

Water is boiled in a pan on a stove at sea level. During 10 min of boiling, it is observed that 200 g of water has evaporated. T

diamong [38]

Answer:

Rate of heat is 45.1 kJ/min

Explanation:

Heat required to evaporate the water is given by

Q = mL

here we know that

$L = 2.25 \times 10^6 J/kg$

now we have

$Q = (0.200)(2.25 \times 10^6 J/kg)$

$Q = 452.1 kJ$

now the power is defined as rate of energy

$P = \frac{Q}{t}$

$P = \frac{452.1 kJ}{10}$

$P = 45.1 kJ/min$

6 0

3 years ago

n an experiment of a simple pendulum, measurements show that the pendulum has length m, mass kg, and period s. Take m/s2 . i. Us

barxatty [35]

Answer:

The answer is " $(1.265 \pm 0.010) \ s \ and \ 0.709 \%$ "

Explanation:

In point i:

$T_{theo}= 2\pi \sqrt{\frac{l}{g}}$

$=2\pi\sqrt{\frac{0.397}{9.8}}\\\\= 1.265 \ s$

If error in the theoretical time period :

$\frac{\Delta T_{theo}}{T_theo} = \frac{1}{2} \frac{\Delta l }{l}\\\\\Delta T_{theo} = 1.265 \times \frac{1}{2} \times \frac{0.006}{0.397}$

$= 0.010 \ s$

$T_{theo} = (1.265 \pm 0.010) \ s$

In point ii:

$\% \ difference = \frac{|T_{exp} -T_{theo}|}{\frac{T_{exp}+T_{theo}}{2}} \times 100$

<h3> $= \frac{1.274 -1.265}{\frac{1.274+1.265}{2}} \times 100\\\\=0.709 \%$ </h3>

5 0

2 years ago

A positive charge, q1, of 5 µC is 3 × 10–2 m west of a positive charge, q2, of 2 µC. What is the magnitude and direction of the

belka [17]

We know, F = 1/4πε * q₁q₂ / r²
Here, q₁ = 5 * 10⁻⁶ C
q₂ = 2 * 10⁻⁶ C
r = 3 * 10⁻² m west

Substitute their values,
F = (9 * 10⁹) (5 * 10⁻⁶) (2 * 10⁻⁶) / (3 * 10⁻²)²

F = 100 N [ East of positive charge ]

Hope this helps!

5 0

3 years ago

Read 2 more answers

A 20 cm tall object is placed in front of a concave mirror with a radius of 31 cm. The distance of the object to the mirror is 9

Aloiza [94]

Answer:

The focal length of the concave mirror is -15.5 cm

Explanation:

Given that,

Height of the object, h = 20 cm

Radius of curvature of the mirror, R = -31 cm (direction is opposite)

Object distance, u = -94 cm

We need to find the focal length of the mirror. The relation between the focal length and the radius of curvature of the mirror is as follows :

R = 2f

f is the focal length

$f=\dfrac{R}{2}$

$f=\dfrac{-31}{2}$

f = -15.5 cm

So, the focal length of the concave mirror is -15.5 cm. Hence, this is the required solution.

4 0

2 years ago