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1 year ago

10

a school bus takes 0.50 hours to reach the school from your home. if the average speed of the bus is 20km/h, what is displacemen

t of the bus the trip

1 answer:

Kazeer [188]1 year ago

8 0

Answer:

10 km

Explanation:

.5 hr * 20 km/hr = 10 km

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Who exerts more pressure? a) A girl of 50 kg, wearing heels with an area of 1 cm2. b) An elephant of 4000 kg with foot area of 2

Mrrafil [7]

Answer:

The girl exerts more pressure.

Explanation:

Pressure can be defined as the force exerted normally or perpendicularly per unit area.

i.e P = F/A

<u>Girls</u>

Area of the heel = 1cm² = 10^(-4) m²

Force = mg = 50 × 10 = 500N

Pressure =

$\frac{500}{10 ^{ - 4} }$

$= 5 \times {10}^{6}$

<u>Elephant</u>

<u>Area</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u>0</u><u>cm</u><u>²</u><u> </u><u>=</u><u> </u><u>2</u><u>.</u><u>5</u><u> </u><u>x</u><u> </u><u>1</u><u>0</u><u>^</u><u>(</u><u>-</u><u>2</u><u>)</u><u>b</u><u> </u><u>m</u><u>²</u>

<u>Force</u><u> </u><u>=</u><u> </u><u>mg</u><u> </u><u>=</u><u> </u><u>4</u><u>0</u><u>0</u><u>0</u><u>0</u><u>N</u>

<u>Pressure</u><u> </u><u>=</u><u> </u>

<u> $\frac{40000}{2.5 \times {10}^{ - 2} }$ </u>

<u> $= 1.6 \times {10}^{6}$ </u>

5 0

3 years ago

An empty water tank has a rectangular base with side lengths of 1.2 m and 2.3 m. The weight of the tank applies a pressure of 35

LekaFEV [45]

Answer:

Weight = 966 Newton.

Explanation:

Given the following data;

Length = 1.2 m

Width = 2.3 m

Pressure = 350 Pa

To find the weight of the tank;

We know that weight is the force of gravity acting on an object multiplied by its mass.

Weight = mg = force

Hence, we would determine the force using the parameters that were given.

But we would first determine the area of the rectangular tank.

Area of rectangle, A = length * width

A = 1.2 * 2.3

A = 2.76 m²

Mathematically, pressure is given by the formula;

Pressure = force/area

Force = pressure * area

Substituting into the formula, we have;

Force = 2.76 * 350

Force = 966 Newton

Therefore, the weight of the tank is 966 Newton.

5 0

3 years ago

Particle A of charge 2.70 10-4 C is at the origin, particle B of charge -6.36 10-4 C is at (4.00 m, 0), and particle C of charge

Serhud [2]

Answer:

FC vector representation

$F_{C} =(19.03i+13.78j)N$

Magnitude of FC

$F_{C}=23.495N$

Vector direction FC

$\beta=35.91$ degrees: angle that forms FC with the horizontal

Explanation:

Conceptual analysis

Because the particle C is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

The directions of the individual forces exerted by qA and qB on qC are shown in the attached figure; The force (FAC) of qA over qC is repulsive because they have equal signs and the force (FBC) of qB over qC is attractive because they have opposite signs.

The FAC force is up in the positive direction and the FBC force forms an α angle with respect to the x axis.

$\alpha = tan^{-1}(\frac{3}{4}) = 36.86$ degrees

To calculate the magnitudes of the forces we apply Coulomb's law:

$F_{AC} = \frac{k*q_{A}*q_{C}}{r_{AC}^2}$ Equation (1): Magnitude of the electric force of the charge qA over the charge qC

$F_{BC} = \frac{k*q_{B}*q_{C}}{r_{BC}^2}$ Equation (2) : Magnitude of the electric force of the charge qB over the charge qC

Known data

$k=8.99*10^9 \frac{N*m^2}{C^2}$

$q_{A}=2.70*10^{-4} C$

$q_{B}=-6.36*10^{-4} C$

$q_{C}=1.04*10^{-4} C$

$r_{AC} =3$

$r_{BC}=\sqrt{4^2+3^2} = 5$

Problem development

In the equations (1) and (2) to calculate FAC Y FBC:

$F_{AC} =8.99*10^9*(2.70*10^{-4}* 1.04*10^{-4})/(3)^2=28.05N$

$F_{BC} =8.99*10^9*(6.36*10^{-4}* 1.04*10^{-4})/(5)^2=23.785N$

Components of the FBC force at x and y:

$F_{BCx}=23.785 *Cos(36.86)=19.03N$

$F_{BCy}=23.785 *Sin(36.86)=14.27N$

Components of the resulting force acting on qC:

$F_{Cx} = F_{ACx}+ F_{BCx}=0+19.03=19.03N$

$F_{Cy} = F_{ACy}+ F_{BCy}=28.05-14.27=13.78N$

FC vector representation

$F_{C} =(19.03i+13.78j)N$

Magnitude of FC

$F_{C}= \sqrt{19.03^2+13.78^2} =23.495N$

Vector direction FC

$\beta = tan^{-1} (\frac{13.78}{19.03})=35.91$ degrees: angle that forms FC with the horizontal

7 0

3 years ago

If a force of 10 N is applied to a 4.5 kg object, then the object will accelerate at 2.22 m/s2. True or False?

zloy xaker [14]

True f=ma
this is newtons second law
simply put the values inthe equation
f=10N
m=4.5
a=f/m

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2 years ago

Can anyone answer this for me

horrorfan [7]

Answer:

answer what broski

Explanation

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2 years ago