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3 years ago

Which is an example of a vector quantity

Physics

1 answer:

swat323 years ago

4 0

Answer:

The answer to your question is C

Explanation:

acceleration is a vector quantity because it has both magnitude and direction

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The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the el

LUCKY_DIMON [66]

Answer:

e.)At twice the distance, the strength of the field is E/4.

Explanation:

The strength of the electric field at a certain distance from a point charge is given by:

$E=k\frac{Q}{r^2}$

where

k is the Coulomb's constant

Q is the charge

r is the distance from the point charge

In this problem, the distance from the point charge is doubled:

r' = 2r

So the new electric field strength is

$E'=k\frac{Q}{(2r)^2}=k \frac{Q}{4 r^2}=\frac{1}{4} (k\frac{Q}{r^2})=\frac{E}{4}$

so, at twice the distance the strength of the field is E/4.

4 0

2 years ago

Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.

Tems11 [23]

Answer:

The range is 35.35 m

Explanation:

Projectile Motion

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and $g=9.8m/s^2$ the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

$\displaystyle d={\frac {v_o^{2}\sin(2\theta )}{g}}$

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

$\displaystyle d={\frac {20^{2}\sin(2\cdot 30^\circ )}{9.8}}$

$\displaystyle d={\frac {400\sin(60^\circ )}{9.8}}$

$d=35.35\ m$

The range is 35.35 m

7 0

2 years ago

What will be the final temperature if a 4.00 g silver ring at 41.0◦C if it gives off 18.0 J of heat to the surroundings? The spe

Eduardwww [97]

Answer:

Final temperature, $T_f=21.85^{\circ}$

Explanation:

Given that,

Mass of silver ring, m = 4 g

Initial temperature, $T_i=41^{\circ}C$

Heat released, Q = -18 J (as heat is released)

Specific heat capacity of silver, $c=0.235\ J/g\ C$

To find,

Final temperature

Solution,

The expression for the specific heat is given by :

$Q=mc\Delta T$

$Q=mc(T_f-T_i)$

$T_f=\dfrac{-Q}{mc}+T_i$

$T_f=\dfrac{-18}{4\times 0.235}+41$

$T_f=21.85^{\circ}$

So, the final temperature of silver is 21.85 degrees Celsius.

5 0

2 years ago

An above ground swimming pool of 30 ft diameter and 5 ft depth is to be filled from a garden hose (smooth interior) of length 10

STALIN [3.7K]

This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

It will take "5.1 hours" to fill the pool.

First, we will use the formula for the dynamic pressure to find out the flow speed of water:

$P=\frac{1}{2}\rho v^2\\\\v=\sqrt{\frac{2P}{\rho}}$

where,

v = flow speed = ?

P = Dynamic Pressure = 55 psi $(\frac{6894.76\ Pa}{1\ psi})$ = 379212 Pa

$\rho$ = density of water = 1000 kg/m³

Therefore,

$v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}$

v = 27.54 m/s

Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

$\frac{V}{t} = Av\\\\t =\frac{V}{Av}$

where,

t = time to fill the pool = ?

A = Area of the mouth of hose = $\frac{\pi (0.015875\ m)^2}{4}$ = 1.98 x 10⁻⁴ m²

V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)

V = $[\frac{\pi (9.144\ m)^2}{4}][1.524\ m]$ = 100.1 m³

Therefore,

$t = \frac{(100.1\ m^3)}{(1.98\ x\ 10^{-4}\ m^2)(27.54\ m/s)}\\\\$

t = 18353.5 s = 305.9 min = 5.1 hours

Learn more about dynamic pressure here:

brainly.com/question/13155610?referrer=searchResults

7 0

2 years ago

A very smart 3-year-old child is given a wagon for her birthday. She refuses to use it. "After all," she says, "Newton's third l

natita [175]

Answer:

Explanation:

She's correct but doesn't mean the wagon cannot put into motion. The force that she applied on the wagon, according to Newton's 2nd law, would have generated an acceleration, which translates into motion. The reaction force the wagon applies on her due to Newton's 3rd law, would not hinder its own motion.

5 0

3 years ago