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kupik [55]
3 years ago
11

A 26-g steel-jacketed bullet is fired with a velocity of 630 m/s toward a steel plate and ricochets along path CD with a velocit

y 500 m/s. Knowing that the bullet leaves a 50-mm scratch on the surface of the plate and assuming that it has an average speed of 600 m/s while in contact with the plate, determine the magnitude and direction of the impulsive force exerted by the plate on the bullet.
Physics
1 answer:
Alecsey [184]3 years ago
5 0

Answer:

  F = - 3.53 10⁵ N

Explanation:

This problem must be solved using the relationship between momentum and the amount of movement.

          I = F t = Δp

To find the time we use that the average speed in the contact is constant (v = 600m / s), let's use the uniform movement ratio

        v = d / t

        t = d / v

Reduce SI system

          m = 26 g ( 1 kg/1000g) = 26 10⁻³ kg

          d = 50 mm ( 1m/ 1000 mm) = 50 10⁻³ m

Let's calculate

         t = 50 10⁻³ / 600

         t = 8.33 10⁻⁵ s

With this value we use the momentum and momentum relationship

        F t = m v - m v₀

As the bullet bounces the speed sign after the crash is negative

       F = m (v-vo) / t

       F = 26 10⁻³ (-500 - 630) / 8.33 10⁻⁵

       F = - 3.53 10⁵ N

The negative sign indicates that the force is exerted against the bullet

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A person walks 20.0° north of east for 3.20 km. How far would she have to walk due north and due east to arrive at the same loca
aivan3 [116]

Answer:

1.09 km, 3 km

Explanation:

displacement, d = 3.20 km at 20° North of east

A vector quantity has two components one along the x axis and the other is along Y axis. the component along X axis is called the horizontal component and it is due east.

The component along y axis is called vertical component and it is due North.

Displacement due North, dy = 3.20 Sin 20° = 1.09 km

Displacement due east, dx = 3.20 Cos 20° = 3 km

8 0
4 years ago
How the forces makes effect on the state of motion of an object ? <br><br> Plz ans fast urgent
Ksju [112]

Answer:

1. It may change the direction of an object in motion.

2. It may cause change in velocity of an object in motion.

Explanation:

1.It may change the direction of an object in motion.

When an object is in motion,an applied force on that object may change its direction.

For example, a sailboat moving eastward, can suddenly change its direction by interaction of a storm wind blowing form the south.

2. It may cause change in velocity of an object in motion .

A force applied to an object in motion can increase or decrease its speed. When the force is applied to the object in motion in the direction of that object, its velocity may increase.

On the other hand, when the force is applied in the opposite direction to the object in motion, its velocity may reduce.

6 0
3 years ago
If the mass of an object is 5 kg and the velocity is 8 m/s, what is the momentum?
topjm [15]

Answer:

40 kg•m/s

Explanation:

momentum (p) = mass (kg) × velocity (m/s)

3 0
3 years ago
This is a change in an object's position. <br> Net Force<br> Gravitational Pull<br> Motion<br> Force
alexira [117]

Answer:

Motion

Explanation:

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3 0
3 years ago
A hydrogen atom contains a single electron that moves in a circular orbit about a single proton. Assume the proton is stationary
riadik2000 [5.3K]

Answer:

Radius between electron and proton= 6.804\times 10^{-10}m

Explanation:

The motion of the electron is carried out in the orbit due to the balancing of the electrostatic force between the proton and the electron and the centripetal force acting on the electron.

The electrostatic force is given as = \frac{kq_1q_2}{r^2}

Where,

k = coulomb's law constant (9×10⁹ N-m²/C²)

q₁ and q₂  = charges = 1.6 × 10⁻¹⁹ C

r = radius between the proton and the electron

Also,

Centripetal force on the moving electron is given as:

=\frac{m_eV^2}{r}

where,

m_e = mass of the electron (9.1 ×10⁻³¹ kg)

V = velocity of the moving electron (given: 6.1 ×10⁵ m/s)

Now equating both the formulas, we have

\frac{kq_1q_2}{r^2} = \frac{m_eV^2}{r}

⇒r = \frac{kq_1q_2}{m_eV^2}

substituting the values in the above equation we get,

r = \frac{9\times 10^{9}\times (1.6\times 10^{-19})^2}{9.1\times 10^{-31}\times (6.1\times 10^5)^2}

⇒r = 6.804\times 10^{-10}m

3 0
3 years ago
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