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elena55 [62]
3 years ago
6

3 نقطة (نقاط)

Physics
1 answer:
Molodets [167]3 years ago
5 0

The work-energy theorem says that the total work done on the block is equal to the difference of its kinetic energies at points B and A. Then the total work done on the block is

W_{\rm total} = K_B - K_A = 4.0\,\mathrm J - 5.0\,\mathrm J = -1.0\,\mathrm J

Friction acts on the block to oppose its motion, so it does negative work on the block, -4.5 J.

The only other force acting on the block as it moves is the force <em>P</em>. Let W_P be the work done by the force <em>P</em>. Then the total work done on the block is

W_{\rm total} = W_P + W_{\rm friction} \iff -1.0\,\mathrm J = W_P - 4.5 \,\mathrm J \implies W_P = \boxed{3.5\,\mathrm J}

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A car (m = 1302 kg) traveling along a road begins accelerating with a constant acceleration of 1.50 m/s2 in the direction of mot
antiseptic1488 [7]

First we will use the concepts of motion kinetics for which the final speed is defined as shown below,

v_f^2=v_i^2+2as

Here,

v_f= Final velocity

v_i= Initial velocity

a = Acceleration

s = Distance

Replacing,

(35)^2 = v_i^2+2(1.5)(392)

v_i = 7m/s

Using the conservation of energy for kinetic energy we have,

KE = \frac{1}{2}mv_i^2

KE = \frac{1}{2}(1302)(7)^2

KE = 31900J

Therefore the kinetic energy of the car is 31900J

5 0
4 years ago
at a certain moment, an object has an amount of 200 of motion energy and 400 of gravitational potential energyThe object is also
Radda [10]

Answer:

twice as much energy

Explanation:

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2 years ago
A. A land speed car can decelerate at 9.8m/s. How long does it take the car to come to a complete stop from a run of 885 km/hr (
Nimfa-mama [501]

Answer:

A. 25.08 s

B. 3082.53 m

C. 3×10⁵ m/s²

Explanation:

A. Determination of the time.

This can be obtained as illustrated below:

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Time (t) =.?

v = u + at

0 = 245.8 + (–9.8 × t)

0 = 245.8 – 9.8t

Collect like terms

0 – 245.8 = – 9.8t

– 245.8 = – 9.8t

Divide both side by –9.8

t = –245.8 / –9.8

t = 25.08 s

Therefore, it will take 25.08 s for the car to come to a complete stop.

B. Determination of the distance travelled by the car.

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Distance (s) =?

v² = u² + 2as

0² = 245.8² + (2 × –9.8 × s)

0 = 60417.64 – 19.6s

Collect like terms

0 – 60417.64 = – 19.6s

– 60417.64 = – 19.6s

Divide both side by –19.6

s = –60417.64 / –19.6

s = 3082.53 m

Thus, the car travelled a distance of 3082.53 m before stopping completely.

C. Determination of the acceleration of the object.

Initial velocity (u) = 0 m/s

Final velocity (v) = 600 m/s

Distance (s) = 0.6 m

Acceleration (a) =?

v² = u² + 2as

600² = 0² + (2 × a × 0.6)

360000 = 0 + 1.2a

360000 = 1.2a

Divide both side by 1.2

a = 360000 / 1.2

a = 300000 = 3×10⁵ m/s²

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Increasing the area of the base will also increase the stability of an object, the bigger the area the more stable the object. Rugby players will stand with their feet well apart if they are standing and expect to be tackled.

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