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2 years ago

What is the acceleration of a 32-kg object if a 6.75-N force is applied to it?

Physics

1 answer:

professor190 [17]2 years ago

5 0

F=MA
*newton 2nd law*
F=6.75N
M=32kg

A= F/M
A=6.75/32
equal
a.0.21m/s2

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At a time t = 2.80 s , a point on the rim of a wheel with a radius of 0.230 m has a tangential speed of 51.0 m/s as the wheel sl

bonufazy [111]

Answer:

Part A) the angular acceleration is α= 44.347 rad/s²

Part B) the angular velocity is 195.13 rad/s

Part C) the angular velocity is 345.913 rad/s

Part D ) the time is t= 7.652 s

Explanation:

Part A) since angular acceleration is related with angular acceleration through:

α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²

Part B) since angular acceleration is related

since

v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s

since

ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s

Part C) at t=0

v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s

ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s

Part D ) since the radial acceleration is related with the velocity through

ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s

therefore

v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s

t= 7.652 s

4 0

2 years ago

A 2.0-kg ball rolls to the right at 3.0 m/s. A 4.0-kg ball rolls to the left at 2.0 m/s . What is the momentum of the system aft

charle [14.2K]

Answer:

Final momentum after a head on collision is -2kgm/s

Explanation:

One ball moves to the right and the other moves opposite and momentum is a vector quantity so that considering the direction

Initial momenta are P₁=2x3=6kgm/s P₂=4x(-2)=-8kgm/s

Final momentum is the vector sum of P(final)= 6-8= -2 kgm/s

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2 years ago

How much heat is evolved when 1234 g of water condenses to a liquid at 100.°c?

lbvjy [14]

During the phase transition vapour --> liquid water, the temperature of the water does not change; the molecules of water release heat and the amounf of heat released is equal to
$Q=mL_e$
where
m is the mass of the water
$L_e$ is the latent heat of evaporation.

For water, the latent heat of evaporation is $L=2257 kJ/kg$ , while the mass of the water is
$m=1234 g=1.234 kg$

so, the amount of heat released in the process is
$Q=mL_e = (1.234 kg)(2257 kJ/kg)=2785 kJ$

6 0

3 years ago

A powerful motorcycle can produce an acceleration of 3.00 m/s2 while traveling at 90.0 km/h. At that speed the forces resisting

nata0808 [166]

Answer:

1185 N

Explanation:

From Newton’s second law of motion,

F=ma where m= mass of motorcycle, a is acceleration of the motorcycle and F=Force

Net force acting on motorcycle $F_{net}$ is given by $F_{net}=F-f$