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3 years ago

I tried looking at formulas, if you know the formula can you list it?

Physics

2 answers:

mash [69]3 years ago

8 0

Answer:

a. -10 m/s²

b. -15000 N

Explanation:

a. First, write which variables are given:

v₀ = 20 m/s

v = 0 m/s

t = 2 s

Next, identify the variable you're looking for.

Find: a

Determine which equation uses these variables. One trick is to find which variable isn't included in the problem, then find the equation that doesn't use that variable. In this case, we don't need displacement Δx. So the equation we should use is:

v = at + v₀

Plug in and solve:

(0 m/s) = a (2 s) + (20 m/s)

a = -10 m/s²

b. Force is mass times acceleration.

F = ma

F = (1500 kg) (-10 m/s²)

F = -15000 N

serg [7]3 years ago

5 0

Answer:y=mx+b

Explanation:

y=mx+b

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If a cannonball is fired horizontally it will not go in a straight line why?

goldfiish [28.3K]

Explanation:

A projectile (Cannon ball) is launched at an angle to the horizontal and rises up to a peak while moving horizontally. When it reaches the peak, the projectile starts to fall.

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1 year ago

Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini

amid [387]

Answer:

Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:

$E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}$

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

$E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}$

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

$V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r$

= $\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]$

∴NOTE: Graph is attached

8 0

3 years ago

What is the frequency of the electromagnetic wave if the period is 1.54x10-15s

pogonyaev

answer✿࿐

I was not able to write it here

so I did it somewhere else and attached the picture

i hope it helps

have a nice day

#Captainpower

3 0

2 years ago

What is the force felt by the 64-kg occupant of the car? Express your answer to two significant figures and include the appropri

Yakvenalex [24]

Answer:

The force is $-1.67\times10^{5}\ N$

Explanation:

Given that,

Mass of car = 64 kg

Suppose, a 1400-kg car that stops from 34 km/h on a distance of 1.7 cm.

We need to calculate the acceleration

Using formula of acceleration

$v^2-u^2=2as$

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value into the formula

$0^2-(34\times\dfrac{5}{18})^2=2\times a\times 1.7\times10^{-2}$

$a=\dfrac{(34\times\dfrac{5}{18})^2}{2\times1.7\times10^{-2}}$

$a=-2623.45\ m/s²$

We need to calculate the force

Using formula of force

$F=ma$

$F=64\times(-2623.45)$

$F=-1.67\times10^{5}\ N$

Negative sign shows the direction of the force is in the direction opposite to the initial velocity.

Hence, The force is $-1.67\times10^{5}\ N$

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3 years ago

A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area

weqwewe [10]

Answer:

W₂= 10000 N

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

Pressure is defined as the force (F) applied per unit area (A)

P=F/A (N/m²)

P1=P2

$\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }$

$F_{2} = \frac{F_{1}*A_{2} }{A_{1}}$ Equation (1)

Data

W₁ = weight sits on the small piston

F₁ = W₁= 500 N

A₁ = 2.0 cm²

A₂ = 40 cm²

Calculation of the weight (W₂) can the large piston support

We replace data in the equation (1)

$F_{2} = \frac{(500)*(40) }{2}$

F₂ = 10000 N

W₂= F₂= 10000 N

6 0

3 years ago