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PilotLPTM [1.2K]
2 years ago
12

A 2 kg object being pulled across the floor with a speed of 10 m/sec is suddenly

Physics
1 answer:
zvonat [6]2 years ago
5 0

Answer:

The frictional force producing this deceleration would have a magnitude of 4\; \rm N.

Explanation:

The velocity of this object changed by \Delta v = (-10\; \rm m\cdot s^{-1}) in \Delta t = 5\; \rm s. The acceleration of this object would be:

\begin{aligned}a &= \frac{\Delta v}{\Delta t} \\ &= \frac{-10\; \rm m\cdot s^{-1}}{5\; \rm s} = -2\; \rm m\cdot s^{-2}\end{aligned}.

Let m denote the mass of this object. By Newton's Second Law of Motion, the net force on this object would be:

\begin{aligned}F &= m \, a \\ &= 2\; \rm kg \times (-2\; \rm m\cdot s^{-2}) \\ &= -4\; \rm N\end{aligned}.

(1\; {\rm kg \cdot m \cdot s^{-2} = 1\; {\rm N}.)

If the floor is level, friction would be the only unbalanced force on this object. Thus, the magnitude of the frictional force on this object would also be 4\; {\rm N}, same as the magnitude of the net force on this object.

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Emmy is standing on a moving sidewalk that moves at +2 m/s. Suddenly, she realizes she might miss her flight, so begins to speed
likoan [24]

Answer:

Refer to the attachment for the diagram.

3.53 m/s.

Explanation:

Acceleration is the first derivative of velocity relative to time. In other words, the acceleration is the same as the slope (gradient) of the velocity-time graph. Let t represents the time in seconds and v the speed in meters-per-second.

For 0 < x \le 1:

  • Initial value of v: \rm 2\;m\cdot s^{-1} at t = 0; Hence the point on the segment: (0, 2).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm 2\; m\cdot s^{-2}.
  • Find the equation of this segment in slope-point form: v - 2 = 2 (t - 0) \implies v = 2t + 2, \quad 0 < t \le 1.

Similarly, for 1 < x \le 2:

  • Initial value of v is the same as the final value of v in the previous equation at t = 1: t = 2t + 2 = 4; Hence the point on the segment: (1, 4).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm 1\; m\cdot s^{-2}.
  • Find the equation of this segment in slope-point form: v - 4 = (t - 1) \implies v = t + 3 \quad 1 < t \le 2.

For 2 < x \le 3:

  • Initial value of v is the same as the final value of v in the previous equation at t = 2: t = t + 3 = 5; Hence the point on the segment: (2, 5).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm 0\; m\cdot s^{-2}. There's no acceleration. In other words, the velocity is constant.
  • Find the equation of this segment in slope-point form: v - 5 = 0 (t - 2) \implies v = 5 \quad 2 < t \le 3.

For 3 < x \le 4:

  • Initial value of v is the same as the final value of v in the previous equation at t = 3: t = 5; Hence the point on the segment: (3, 5).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm -3\; m\cdot s^{-2}. In other words, the velocity is decreasing.
  • Find the equation of this segment in slope-point form: v - 5 = -3 (t - 3) \implies v = -3t + 14 \quad 3 < t \le 4.

For 4 < x \le 5:

  • Initial value of v is the same as the final value of v in the previous equation at t = 4: t = -3t + 14; Hence the point on the segment: (4, 2).
  • Slope of the velocity-time graph is the same as acceleration during that period of time: \rm 0\; m\cdot s^{-2}. In other words, the velocity is once again constant.
  • Find the equation of this segment in slope-point form: v - 2 = 0 (t - 4) \implies v = 2\quad 4 < t \le 5.

t = \rm 3.49\;s is in the interval 3 < x \le 4. Apply the equation for that interval: v = -3t +14 = \rm 3.53\; m \cdot s^{-1}.

4 0
3 years ago
PLS ANSWER ASAP
TiliK225 [7]

Answer: Answer is D

I took the test little while back.

3 0
3 years ago
A projectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. At the highest poin
tekilochka [14]

Answer:

a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b)  Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

       sin 40 = voy / v₀

       v_{oy} = v₀ sin 40

       v_{oy} = 50.0 sin40

       v_{oy} = 32.14 m / s

       cos 40 = v₀ₓ / V₀

       v₀ₓ = v₀ cos 40

       v₀ₓ = 50.0 cos 40

       v₀ₓ = 38.3 m / s

Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis

Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and v_{y} = -10.0 m / s

Fragment mass 2 m₂ = 0.7 kg moves in the x direction

Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

     p₀ₓ = m v₀ₓ

Y Axis y

    p_{oy} = 0

After the break

X axis

   p_{fx} = m₂ v₂

Axis y

     p_{fy} = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis  

     0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

     v₃ = - m₁ v₁ / m₃

     v₃ = - (-10) 1 / 0.3

     v₃ = 33.3 m / s

X axis

     M v₀ₓ = m₂ v₂

     v₂ = v₀ₓ M / m₂

     v₂ = 38.3  2 / 0.7

     v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

       v_{fy}² = v_{oy}² - 2 gy

       0 =  v_{oy}²-2gy

       y =  v_{oy}² / 2g

       y = 32.14² / 2 9.8

       y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

      v_{f}² =  v_{y}² - 2 g y₂

      0 =  v_{y}² - 2 g y₂

     y₂ =  v_{y}² / 2g

     y₂ = 33.3²/2 9.8

     y₂ = 56.58 m

Total body height is

      Y = y + y₂

      Y = 52.7 + 56.58

     Y = 109.3 m

8 0
2 years ago
The drawing shows a plot of the output emf of a generator as a function of time t. The coil of this device has a cross-sectional
adell [148]

This question is incomplete, the missing image uploaded along this answer below;

Answer:

a) frequency is 2.381 Hz

b) the angular frequency/speed is 14.96 rad/s

c) the magnitude of the magnetic field is 0.5199 T

Explanation:

Given that;

cross sectional Area A = 0.018 m²

Number of turns N = 200

from the diagram. maximum time T = 0.42 sec

a) the frequency f of the generator in hertz

frequency = 1 / T

we substitute

frequency = 1 / 0.42 = 2.381 Hz

Therefore, frequency is 2.381 Hz

b)  the angular frequency in rad/s

angular speed ω = 2πf

we substitute

ω = 2π × 2.381

ω = 14.96 rad/s

Therefore, the angular frequency/speed is 14.96 rad/s

c) the magnitude of the magnetic field.

to determine the magnitude of the magnetic field, we use the following expression;

e = NBAω

from the diagram, e = 28.0 V

so we substitute

28.0 V = 200 × B × 0.018 × 14.96

28 = 53.856B

B = 28 / 53.856

B = 0.5199 T

Therefore, the magnitude of the magnetic field is 0.5199 T

6 0
3 years ago
1 Point
irakobra [83]

Answer:

SECOND LAW OF NEWTON

Explanation:

When the rocket fires the engines the gases leave at high speed and collide with the space station, transferring an impulse given by the expression

                I = F t = Δp

As we can see this expression is a form of Newton's second law

           F = m a

           a = dv / dt

           F = m dv / dt

           F dt = m dv

           p = mv

           F dt = dp

Therefore the station moves through the SECOND LAW OF NEWTON

7 0
3 years ago
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