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REY [17]
3 years ago
7

A car has an initial velocity of 50 m/s and a constant

Physics
1 answer:
Rasek [7]3 years ago
3 0
Vf=vi+at
Vf= (50m/s)+ (5m/s2)(3s)
Vf=65m/s
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Identify two ways in which friction is not useful in science​
kiruha [24]

Answer:

  • Friction between brakes and the wheels of a car or bicycle helps the vehicles slow down.
  • If you do not lubricate your bike regularly with oil, the friction in the chain and axles increases.
8 0
3 years ago
A mass suspended from a spring is oscillating up and down as indicated. Consider the following possibilities. A At some point du
dangina [55]

A mass suspended from a spring is oscillating up and down, (as stated but not indicated).

A). At some point during the oscillation the mass has zero velocity but its acceleration is non-zero (can be either positive or negative).  <em>Yes. </em> This statement is true at the top and bottom ends of the motion.

B). At some point during the oscillation the mass has zero velocity and zero acceleration.  No.  If the mass is bouncing, this is never true.  It only happens if the mass is hanging motionless on the spring.

C). At some point during the oscillation the mass has non-zero velocity (can be either positive or negative) but has zero acceleration.  <em>Yes.</em>  This is true as the bouncing mass passes through the "zero point" ... the point where the upward force of the stretched spring is equal to the weight of the mass.  At that instant, the vertical forces on the mass are balanced, and the net vertical force is zero ... so there's no acceleration at that instant, because (as Newton informed us), A = F/m .  

D). At all points during the oscillation the mass has non-zero velocity and has nonzero acceleration (either can be positive or negative).  No.  This can only happen if the mass is hanging lifeless from the spring.  If it's bouncing, then It has zero velocity at the top and bottom extremes ... where acceleration is maximum ... and maximum velocity at the center of the swing ... where acceleration is zero.  

7 0
3 years ago
A plane flies from base camp to lake a, 205 km away in the direction 20.0° north of east. after dropping off supplies it flies
miss Akunina [59]
Draw a diagram to illustrate the problem as shown in the figure below.

Camp A  is 20° north of east from the camp, therefore
m∠CAB = 80°
where C => base camp.

Let d = distance from lake B to the base camp, at x° west of south.
Apply the Law of Cosines to determine d.
d² = 205² + 175² - 2*205*175*cos80⁰
     = 6.0191 x 10⁴
d = 245.338 km

Apply the Law of Sines to obtain
\frac{sin(x+30)}{205}= \frac{sin80^{o}}{245.338}\\ \\ sin(x+30)=( \frac{205}{245.338})sin80^{o}=   0.8229
x+30 = sin⁻¹ 0.8229 = 55.4°
x = 25.4°

Answer:
The distance from lake B to base camp is 243.3km (nearest tenth).
The direction is 25.4° west of south.

7 0
3 years ago
If the mass of an object is 300kg and the applied force is 3900N, calculate the acceleration
Scilla [17]
F = ma
3900 = 300a
a = 3900÷300
a = 13m/s/s
5 0
3 years ago
A CD has an initial angular speed of 600 rpm. If the disc stop rotating after 4 seconds, what is its angular acceleration?
Elza [17]

Answer:

-15.708 rad/s^2

Explanation:

First, let us covert everything to the same unit. For me, I find dealing with radians/sec more intuitive, but you can solve it in rpm. We are told that the initial angular speed is 600 rpm and after 4 seconds it stops. Let's convert 600 rpm into radians/sec. To do this, multiply by 2*pi/60. This gives 62.83 rad/s. Now let's review our info:

\omega_i = 600rpm = 62.83rad/s\\\omega_f = 0\\t = 4s\\\alpha = ?

Now we look up angular kinematics equations and the equation that has these parameters is

\omega=\omega_0+\alpha t

Substitute our values in:

\omega=\omega_0+\alpha t\\0=62.83\frac{rad}{s}+\alpha *(4s)\\\alpha = -15.708\frac{rad}{s^2}

5 0
3 years ago
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