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3 years ago

How do you find the volume of a rectangular brick?

Physics

2 answers:

Sunny_sXe [5.5K]3 years ago

5 0

LxWxH always remembered that

timurjin [86]3 years ago

3 0

Answer:

Multiply the length, width and height of the brick to calculate its volume.

Explanation:

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What will happen if an object or liquid with a density less than 1.0 g/cm3 is placed into water; do the same for an object or li

Anna71 [15]

If the object has a lower density than water, it will float.

If the object has a higher density than water, it will sink.

3 0

3 years ago

Near the surface of Earth an electric field points radially downward and has a magnitude of approximately 100 N/C. What charge (

pentagon [3]

Answer:

$q=2.997\times 10^{-4}$ C

Sign-Negative

Explanation:

We are given that

Electric field = $E=100NC^{-1}$ (Radially downward)

Acceleration= $0.19 ms^{-2}$ (Upward)

Mass of charge=3 g= $3\times 10^{-3}$ kg

1kg=1000g

We have to find the magnitude and sign of charge would have to be placed on a penny .

By newton's second law

$\sum F_y=ma$

$\sum F_y=qE-mg$

Substitute the values then we get

$qE-mg=ma$

Substitute the values then we get

$q(100)-3\times 10^{-3}(9.8)=3\times 10^{-3}(0.19)$

$100q-29.4\times 10^{-3}=0.57\times 10^{-3}$

$100q=0.57\times 10^{-3}+29.4\times 10^{-3}=29.97\times 10^{-3}$

$q=\frac{29.97\times 10^{-3}}{100}$

$q=2.997\times 10^{-4}$ C

Sign of charge =Negative

Because electric force acting in opposite direction of electric field therefore,charge on penny will be negative.

4 0

4 years ago

A hiker walks due east for a distance of 25.5 km from her base camp. On the second day, she walks 41.0 km northwest till she dis

GarryVolchara [31]

Resultant displacement is 29.2 km at $83.1^{\circ}$ north of west

Explanation:

To solve the problem, we have to use the rules of vector addition, resolving first each vector along the x- and y- direction.

Taking east as positive x direction and north as positive y- direction, we have:

- First displacement is 25.5 km east, therefore its components are

$A_x = 25.5 km\\A_y = 0 km$

- Second displacement is 41.0 km northwest, so its components are

$B_x = (41.0)cos(135^{\circ})=-29.0 km\\B_y =(41.0)sin(135^{\circ})=29.0 km$

So, the components of the resultant displacement are

$R_x=A_x+B_x=25.5+(-29.0)=-3.5 km\\R_y=A_y+B_y=0+29.0=29.0 km$

And so, the magnitude is calculated using Pythagorean's theorem:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(-3.5)^2+(29.0)^2}=29.2 km$

And the direction is given by

$\theta=tan^{-1}(\frac{R_y}{|R_x|})=tan^{-1}(\frac{29.0}{3.5})=83.1^{\circ}$

Where the angle is measured from the west direction, since Rx is negative.

Learn more about displacement:

brainly.com/question/3969582

#LearnwithBrainly

3 0

3 years ago

A small, single engine airplane is about to take off. The airplane becomes airborne, when its speed reaches 161.0 kmph. The cond

lukranit [14]

Answer:

384.6 m

Explanation:

The length of the runway airport should be long enough to accommodate the aircraft during its acceleration from rest to 161 km/h at rate of 2.6 m/s. We can use the following equation of motion to solve for this:

$v^2 - v_0^2 = 2a\Delta s$

where v0 = 0 m/s is the initial velocity of the airplane when it start accelerating, v = 161 km/h = 161*1000*(1/60)(1/60) = 44.72 m/s is the airborn speed, a = 2.6 m/s2 is the acceleration, and $\Delta s$ is the distance of the runway, which we care looking for