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3 years ago

NEED HELP ASAP

Physics

2 answers:

maksim [4K]3 years ago

4 0

the above three pictures may help you

go through the attachments

Softa [21]3 years ago

3 0

Answer:

Block A velocity is 23.33 m/s and the collission is not elastic.

Explanation:

a) m1v1 + m2v2 = m1v1' + m2v2'

Plug in givens

90+60=3v1'+80

solve for v1'= 23.33m/s

b) Find the initial and final kinetic energy of Block B

Ki= 1/2(4)(15)^2 + 1/2(3)(30)^2 = 1800 J

Kf= 1/2(4)(20)^2 + 1/2(3)*(23.33)^2= 1616.433J

Since Ki does not equal Kf the collision is not elastic

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Two cars are traveling at the same speed of 27 m/s on a curve thathas a radius of 120 m. Car A has a mass of 1100 kg and car B h

Lera25 [3.4K]

Answer:

$a_{cA} = 6.075$ m/s²

$a_{cB} = 6.075$ m/s²

$F_{cA} = 6682.5 N$

$F_{cB} = 9720 N$

Explanation:

Normal or centripetal acceleration measures change in speed direction over time. Its expression is given by:

$a_{c} = \frac{v^{2} }{r}$ Formula 1

Where:

$a_{c}$ : Is the normal or centripetal acceleration of the body ( m/s²)

v: It is the magnitude of the tangential velocity of the body at the given point

.(m/s)

r: It is the radius of curvature. (m)

Newton's second law:

∑F = m*a Formula ( 2)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

Data

$v_{A} = 27 \frac{m}{s}$

$v_{B} = 27 \frac{m}{s}$

$m_{A} = 1100 kg$

$m_{B} = 1600 kg$

r= 120 m

Problem development

We replace data in formula (1) to calculate centripetal acceleration:

$a_{cA} = \frac{(27)^{2} }{120}$

$a_{cA} = 6.075$ m/s²

$a_{cB} = \frac{(27)^{2} }{120}$

$a_{cB} = 6.075$ m/s²

We replace data in formula (2) to calculate centripetal force Fc) :

$F_{cA} = m_{A} *a_{cA} = 1100kg*6.075\frac{m}{s^{2} }$

$F_{cA} = 6682.5 N$

$F_{cB} = m_{B} *a_{cB} = 1600kg*6.075\frac{m}{s^{2} }$

$F_{cB} = 9720 N$

4 0

4 years ago

Place any object (textbook, pen, or eraser) on a floor what happens then

Levart [38]

Answer:

The force of gravity on earth, no matter the object is approximately 9.8 m/ s2 . The reason the crumpled paper hits the ground first is because of air resistance. A crumpled piece of paper has less surface area than an piece of paper that is not crumpled. More surface area means more air resistance.

7 0

2 years ago

A 2.81 μF capacitor is charged to 1220 V and a 6.61 μF capacitor is charged to 560 V. These capacitors are then disconnected fro

fenix001 [56]

Answer:

756.88 Volts will be the potential difference across each capacitor.

Explanation:

$Q=C\times V$

Q = Charge on capacitor

C = Capacitance

V = Voltage across capacitor

Capacitance of first capacitor = $C_1=2.81 \mu F=2.81\times 10^{-6} F$

Charge of first capacitor = $Q_1$

Voltage across first capacitor = $V_1=1220 V$

$Q_1=C_1V_1$

$Q_1=2.81\times 10^{-6} F\times 1220 V=0.0034282 C$

Capacitance of first capacitor = $C_2=6.61\mu F=6.61\times 10^{-6} F$

Charge of second capacitor = $Q_2$

Voltage across first capacitor = $V_2=560 V$

$Q_2=C_2V_2$

$Q_1=6.61\times 10^{-6} F\times 560 V=0.0037016 C$

Both the capacitors are disconnected and positive plates are now connected to each other and the negative plates are connected to each other. These capacitors are connected in parallel combination.

Total charge = Q

$Q =Q_1+Q_2=0.0034282 C+ 0.0037016 C=0.0071298 C$

Total capacitance in parallel combination:

$C_p=C_1+C_2=2.81\times 10^{-6} F+6.61\times 10^{-6} F=9.42\times 10^{-6} F$

Potential across both capacitors = V

$V=\frac{Q}{C}=\frac{0.0071298 C}{9.42\times 10^{-6} F}=756.88 V$

756.88 Volts will be the potential difference across each capacitor.

8 0

3 years ago

___________________ and ___________________________ are two main types of waves.

iris [78.8K]

Answer:

its A

Explanation:

6 0

4 years ago

A lump of clay whose rest mass is 4 kg is travelling at three-fifths the speed of light when it collides head-on with an identic

Harman [31]

Answer:

mass of the composite lump is 10 kg

Explanation:

given data

mass = 4 kg

to find out

mass of composite lump

solution

we know energy is conserved so

so m1 = m2 = m0 that is 4kg

and

E(1) release+ E(2) release = E(1,2) rest

so γ(1)m(1)c² + γ(2)m(2)c² = Mc² ..........................1

that why here

|v(1)| = |v(2)| = 3/5 c ......................2

and

γ = 1 / √(1 − v²/c²) .......................3

put here v = 3 and c is 5

γ = 1 /√(1 − 9/25)

γ = 5/4

γ(1) = γ(2) = γ = 5/4

so from equation 1

γ(1)m(1)c² + γ(2)m(2)c² = Mc²

M = 2γm0

M = 2(5/4 )(4)

M = 10 kg

so mass of the composite lump is 10 kg

7 0

4 years ago

~~~~NEED HELP ASAP~~~~

NEED HELP ASAP