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Charra [1.4K]
3 years ago
14

1.A motorcycle’s velocity at the top of the hill is 11.0 m/s. 4.0 seconds later it reaches the bottom of the hill with a velocit

y of 20.0 m/s. What is the acceleration of the motorcycle?
2.A motorcycle’s velocity at the top of the hill is 11.0 m/s. 4.0 seconds later it reaches the bottom of the hill with a velocity of 20.0 m/s. What is the acceleration of the motorcycle?
Physics
1 answer:
daser333 [38]3 years ago
6 0

Answer:

1) as far as I remember

Let's take 20 as vf (final velocity) and 11 as (initial velocity) and 4 as time

So we would use this formula a=vf-vi/t

So 20-11/4

Asnwer 2.25

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when striking, the pike, a predatory fish, can accelerate from rest to a speed of 4.0 m/sm/s in 0.15 ss
Alina [70]

The distance covered will  be 0.301m.

firstly we will new to calculate acceleration (a).

we can do so using newtons first law of motion.

v=u + a × t

where

a is the acceleration

t is the time

from here we get

a=(v-u)/t

a=(4-0)/0.15

a=26.67m/s²

using newtons second law of motion

s=u.t+1/2at²

where s is distance travelled in t seconds.

we get,

s=0.301m

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brainly.com/question/129361

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6 0
2 years ago
If the satellite has a mass of 3900 kg , a radius of 4.3 m , and the rockets each add a mass of 210 kg , what is the required st
ehidna [41]

Answer:

The required steady force of each rocket is 28.79 N

Explanation:

mass of the satellite, M=3900 kg

radius, r=4.3 m

mass of rocket, m=210 kg

time, t=5.4 min

Moment of Inertia:

I = 1/2 (Mr^2) + 4mr^2

I = 1/2 ( 3900* (4.3)^2) + 4 (210)*(4.3)^2

I = 51587.1 kg m^2

the angular acceleration is:

a= w/t

here w= 2*π*30

so,

a= 2*π*30 / 5.4* 3600

a=0.0096 rad/ s^2

the Torque becomes:

T=I*a = 4r*F

( 51587.1 )*(0.0096) = 4*4.3* F

F= 28.79 N

the required steady force of each rocket is 28.79 N

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7 0
1 year ago
A granite monument has a volume of 25,365.4 cm3. The density of granite is 2.7 g/cm3. Use this information to calculate the mass
Nana76 [90]
V = 25,364.4 cm^3   Is volumer = 2.7g/cm^3      Is density
To calculate mass you use formula:m= V*rTo avoid remembering this formula you can see the type of unit on each given variable. We can see that we have g/cm^3 and cm^3. If we multiply them, we negate cm^3 and cm^3 and we are left with g which is unit for mass.
the answer is :
m = 68,486,6 g   
6 0
3 years ago
Jkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
Marrrta [24]
Loooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooool
7 0
3 years ago
A horizontal uniform bar of mass 3 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the
kirill115 [55]

Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂  : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F  : Force ( N)

d  : Horizontal distance of the force to the point 2 ( N*m

Data

mb = 3 kg : mass of the  bar

mm = 1.5 kg :  mass of the  monkey

L = 3m : lengt of the bar

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the bar

T₁ : Tension in string 1 (vertical upward)

T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

Wm: Weihgt of the monkey  (vertical downward)

Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances  from forces the point 2

d₁₂ = (3-0.6) m = 2.4m  : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium  of moments at the point  2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

5 0
3 years ago
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