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Olin [163]
3 years ago
15

A car moved 60 meters west in 2 hours. What is its average velocity?

Physics
1 answer:
Fofino [41]3 years ago
5 0

Explanation: Velocity is the displacement of an object during a specific unit of time. Two measurements are needed to determine velocity. Displacement and time. Displacement includes a direction, so velocity also includes a direction. Speed with direction. Velocity can be an average velocity or an instantaneous velocity. Units for velocity are the same as for speed: m/s, km/h, and mph. Delta x(Δx) is the symbol used for displacement. Delta (Δ) means to "change in." Δx means to "change in position." Δx is calculated by final position minus initial position. Velocity formula: → v=Δx/t as a fraction.

v=Δx/t

v=\frac{xf-xi}{t}= \frac{60m}{2}=30m

<em><u>Final answer is 30.</u></em>

Hope this helps!

Thanks!

Have a great day!

-Charlie

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ann drove to the store 10 km north of her house and then drove to the library which is 5 km south of the store
lbvjy [14]
5 km North

Your welcome
4 0
3 years ago
A bungee cord can stretch, but it is never compressed. When the distance between the two ends of the cord is less than its unstr
Ksju [112]

Answer:

Explanation:

Given that

g=9.8m/s²

The spring constant is

k=50N/m

The length of the bungee cord is

Lo=32m

Height of bridge which one end of the bungee is tied is 91m

A steel ball of mass 92kg is attached to the other end of the bungee.

The potential energy(Us) of the steel ball before dropped from the bridge is given as

P.E= mgh

P.E= 92×9.8×91

P.E= 82045.6 J

Us= 82045.6 J

Potential energy)(Uc) of the cord is given as

Uc= ½ke²

Where 'e' is the extension

Then the extension is final height extended by cord minus height of cord

e=hf - hi

e=hf - 32

Uc= ½×50×(hf-32)²

Uc=25(hf-32)²

Using conservation of energy,

Then,

The potential energy of free fall equals the potential energy in string

Uc=Us

25(hf-32)²=82045.6

(hf-32)² = 82045.6/25

(hf-32)²=3281.825

Take square root of both sides

√(hf-32)²=√(3281.825)

hf-32=57.29

hf=57.29+32

hf=89.29m

We neglect the negative sign of the root because the string cannot compressed

3 0
3 years ago
Blind spots of large vehicles are called
AlladinOne [14]
<span>A-pillar blind spot. A blind spot in a vehicle is an area around the vehicle that cannot be directly observed by the driver while at the controls, under existing circumstances. Blind spots exist in a wide range of vehicles: cars, trucks, motorboats, sailboats. and aircraft.</span>
6 0
3 years ago
Read 2 more answers
You have 4.9 g of an unknown substance. To identify the substance, you decide to measure its specific heat and find that it requ
kifflom [539]

The unknown substance can be lithium, which has a specific heat capacity of approximately 3.5 J/gK.

Explanation:

When heat energy is supplied to a certain substance, the temperature of the substance increases according to the equation:

Q=mC_s \Delta T

where

Q is the amount of energy supplied

m is the mass of the sample

C_s is the specific heat capacity of the substance

\Delta T is the change in temperature

In this problem, we have

m = 4.9 g is the mass

Q = 668.85 J is the specific heat capacity

\Delta T = 39 K is the change in temperature

Solving for C_s, we find the specific heat capacity of the substance:

C_s = \frac{Q}{m\Delta T}=\frac{668.85}{(4.9)(39)}=3.5 J/gK

Looking at tables of specific heat capacity, we can see that the unknown substance can be lithium, which has a specific heat capacity of approximately 3.5 J/gK.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

3 0
3 years ago
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
3 years ago
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