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Doss [256]
3 years ago
13

You push a heavy crate out of a carpeted room and down a hallway with a waxed linoleum floor. While pushing the crate 2.5 m out

of the room you exert a force of 385 N; while pushing it 10 m down the hallway you exert a force of 130 N. To slow it down you exert a force of 350 N through a distance of 2.0 m, opposite to the motion. How much work do you do in all?
Physics
1 answer:
Alinara [238K]3 years ago
3 0

Answer:

W = 1562.5 J

Explanation:

Path 1:

W₁ = F₁*d₁ = 385 N * 2.5 m = 962.5 J

Path 2:

W₂ = F₂*d₂ = 130 N * 10 m = 1300 J

Path 3:

W₃ = F₃*d₃ = (-350 N) * 2 m = - 700 J  (opposite to the motion)

We get

W = W₁ + W₂ + W₃ = 962.5 J + 1300 J + (- 700 J) = 1562.5 J

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tester [92]
The temperature of the lithosphere is around 300<span>°C</span> - 500<span>°<span>C</span></span>


5 0
3 years ago
Choose true or false for each statement regarding a parallel plate capacitor.. The voltage of a disconnected charged capacitor i
OlgaM077 [116]

Answer:

Explanation:

The voltage of a disconnected charged capacitor increases when the plate area is decreased.

When plate area decreases , capacitance C decreases , but charge Q remains constant .

Q = C V where C is capacitance and V is voltage .

when C decreases , V increases for keeping Q constant .

So the statement is true.

The electric field is dependent on the charge density on the plates.

This statement is true .

The voltage of a connected charged capacitor remains the same when the plate area is decreased .

For a connected capacitor , V or voltage is constant which is equal to voltage of charging battery .

So the statement is true .

3 0
3 years ago
Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time–when they\'re not sleep
Anit [1.1K]

Answer:

mice total momentum (-0.000250, 0.00639) Kg m

Explanation:

To calculate the moment of the mice we must multiply their mass by their velocities, remember that the moment is a vector quantity, so we use the components of velocity

mouse 1

  m1 = 0.0225 Kg

  V1 = (0.869, -0.283) m / s

 

  Px = m Vx

  Px1 = 0.0225 0.869

  Px1 = 0.01955 Kg m

  Py = m Vy

  Py1 = 0.0225 (-0.283)

  Py1 = -0.006368 Kg m

  P1 = (0.0196, -0.00637) Kg m

Mouse 2

 m2 = 0.0223 Kg

 Px2 = 0.0223 (-0.883) = -0.0196 Kg m

 Py2 = 0.0223 (-0.253) = -0.00564 Kg m

 P2 = (-0.0196, -0.00564) Kg m

Mouse 3

 m3 = 0.0197

 Px3 = 0.0197 0.345 = 0.00680 Kg m

 Py3 = 0.0197 0.803 = 0.0158 Kg m

 P3 = (0.00680, 0.0158) Kg m

Mouse4

  m4 = 0.0127 Kg

  Px4 = 0.0127 (-0.555) = -0.00705 Kg m

  Py4 = 0.0127 0.205 = 0.00260 Kg m

  P4 = (-0.00705, 0.00260) Kg m

To find the total momentum we must add each component of the individual moments

   Px = Px1 + Px2 + Px3 + Px4  

   Py = py1 + Py2 + Py3 + Py4

   Px = 0.0196 -0.0196 +0.00680 -0.00705

   Px = -0,000250 Kg m

   Py = -0.00637 -0.00564 +0.0158 +0.00260

   Py = 0.00639 Kg m

   P = (-0.000250, 0.00639) Kg m

7 0
3 years ago
A proton moves at a constant velocity of 50m/s along the x axis ,through crossed electric and magnetic fields .The magnetic fiel
shusha [124]

Answer:

42 I THE CORRECT ANSWER

Explanation:

BECAUSE I KNOW

6 0
3 years ago
A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.
denpristay [2]

Answer:

Approximately 5.5\; \rm N, assuming that the volume of these two charged objects is negligible.

Explanation:

Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.

Let q_1 and q_2 denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, q_1 = 20 \times 10^{-6}\; \rm C  and q_2 = 15 \times 10^{-6}\; \rm C.

Let r denote the distance between these two point charges. In this question, r = 0.7\; \rm m.

Let k denote the Coulomb constant. In standard units, k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}.

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}.

Substitute in the values and evaluate:

\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}.

8 0
3 years ago
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