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3 years ago

How does friction slow things down?

2 answers:

8 0

Friction slows things down because the tiny rough bumps on all surfaces hook on each other and slow down the object and produce heat

Ymorist [56]3 years ago

5 0

Friction slows things down because the force that is created through the interaction between two objects begins to decrease the acceleration and velocity of the object. Therefore, the object begins to slow down.

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In 1967, new zealander burt munro set the world record for an indian motorcycle, on the bonneville salt flats in utah, with a ma

Fed [463]

Draw a velocity-time diagram as shown below.

Because a velocity of 26.82 m/s is attained in 4.00 s from rest, the average acceleration is
a = 26.82/4 = 6.705 m/s²
The time required to reach maximum velocity of 82.1 m/s is
t₁ = (82.1 m/s)/(6.705 m/s²) = 12.2446 s

The distance traveled during the acceleration phase is
s₁ = (1/2)at₁²
= (1/2)*(6.705 m/s²)*(12.2446 s)²
= 502.64 m

Answer:
The time required to reach maximum speed is 12.245 s
The distance traveled during the acceleration phase is 502.6 m

3 0

3 years ago

Three carts of masses 4.0 kg, 10kg, and 3.0 kg move on a frictionless track with speeds of v1 = 5.0m/s, v2=3.0m/s, and v3=-3.6 m

gogolik [260]

2.24 m/s is the calculated velocity.

Initial velocity (u) squared plus two times the acceleration (a) times the displacement equals final velocity (v) squared (s). Final velocity (v) is equal to the square root of initial velocity (u) squared plus two times the acceleration (a) times displacement when v is the variable being solved for (s).

The cart's masses and speeds are known.

M1 = 4.00 kg, M2 = 10.0 kg, M3 = 3.00 kg, etc.

v1 = 5.00 m/s = 5.00 m/s, v2 = 3.00 m/s = 3.00 m/s, v3 = -4.00 m/s = 4.00 m/s, and m1v1+m2v2+m3v3 = (m1+m2+m3) v=d frac m 1v 1+m 2v 2+m 3v 3, where (m1+m2 + m3) is the product of (v1 v 1+m2v2+m3v3).

"m 1+m 2+m 3" is equivalent to "m 1+m2+m3/m1v 1+m2v2 +m3v3"

the three carts' final velocities are calculated as follows: v=d frac

{4.00kg\sdot5.00m/s+10.0kg\sdot3.00m/s-3.00kg\sdot4.00m/s} 4.24m/s = 4.50kg+10.0kg+3.00kg vs. 4.50kg+10.0kg+3.00kg

5.00m/s/4.00kg/5.00m/s+10.0m/s/3.00m/s/4.00m/s =2.24m/s.

2.24 m/s is the calculated final velocity.

Learn more about velocity here-

brainly.com/question/18084516

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7 0

2 years ago

Which statement is true about an airplane wing during flight?

allochka39001 [22]

C. air above the wing travels faster than air below the wing.

3 0

3 years ago

Read 2 more answers

Una persona de 80kg intenta bajar de peso subiendo una montaña para quemar el equilante a una rebanada de pastel de chocolate (7

SCORPION-xisa [38]

Answer:

Explanation:

An 80kg person tries to lose weight by climbing a mountain to burn the equivalent of a large slice of a rich chocolate cake (700kcl). How high should you climb?

Given that,

A person of mass

M = 80kg

The person is climbing a stair case and he burn an energy of 700 kcl

We know that 1kcl = 4184J

Then, 700kcl = 700 × 4184J

700kcl = 2.9288 × 10^6 J

We want to find height of stairs the person will climb when he Loss 700kcl. The energy he burns will be converted to the potential energy

P.E is calculated using

P.E = mgh

Where

m is mass in kg

g is gravitational constant g = 9.81m/s2

h is the height of stairs he climb in metre

Then,

P.E = energy burn

mgh = 700kcl

80 × 9.81 × h = 2.9288 × 10^6

h = (2.9288 × 10^6) / (80 × 9.81)

h = 3731.91m

So, the person will climb 3731.91m before he losses 700kcl

In espanyol

Dado que,

Una persona de masa

M = 80 kg

La persona está subiendo una escalera y quema una energía de 700 kcl.

Sabemos que 1kcl = 4184J

Entonces, 700kcl = 700 × 4184J

700kcl = 2.9288 × 10 ^ 6 J

Queremos encontrar la altura de las escaleras que la persona subirá cuando pierda 700kcl. La energía que quema se convertirá en energía potencial.

P.E se calcula utilizando

P.E = mgh

Dónde

m es masa en kg

g es constante gravitacional g = 9.81m / s2

h es la altura de las escaleras que sube en metros

Entonces,

P.E = quema de energía

mgh = 700kcl

80 × 9.81 × h = 2.9288 × 10 ^ 6

h = (2.9288 × 10 ^ 6) / (80 × 9.81)

h = 3731.91m

Entonces, la persona subirá 3731.91m antes de perder 700kcl

5 0

3 years ago

When hiking at altitude, the air pressure drops to 825,000Pa. Solve the force that site exerts against the side of a rectangular

OLga [1]

Answer:

The force is $1237500N$

Explanation:

We can apply the following expression

$pressure=\frac{Force}{Area}$

To find the area of rectangular tent wall:

$Area=length *width\\Area=2.00m*0.750m\\Area=1.5m^{2}$

Now to find the force:

$pressure=825,000pa$

$pressure=\frac{Force}{Area}$

$Force=pressure*Area\\Force=825000pa*1.5m^{2} \\Force=1237500N$

7 0

3 years ago