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umka21 [38]
2 years ago
10

A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes

and the car comes to a rest uniformly in a distance of 160 m. What are the magnitude and direction of the net force applied to the car to bring it to rest?
Physics
1 answer:
Ludmilka [50]2 years ago
7 0

Answer:

Force applied to stop the car = 1,250 N

Explanation:

Given:

Mass of car (M) = 1,000 kg

Initial velocity (U) = 20 m/s

Final velocity (V) = 0 m/s

Distance (S) = 160 m

Find:

Force applied to stop the car.

Computation:

v^2 = u^2 + 2as\\\\0^2=20^2+2(a)(160)\\\\0=400+320(a)\\\\Acceleration = a = -1.25m/s^2\\\\Force = ma \\\\Force= 1,000(1.25)\\\\Force = 1,250 N

Force applied to stop the car = 1,250 N

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The repeating pattern of a minerals particles in a solid is called
Klio2033 [76]

Answer:

Crystal structure

Explanation:

The repeated pattern of similar particles in a material is called crystal. Crystal structure is the largest constituent unit of a solid matter.

The fundamental identity of a crystal structure is a unit cell that is formed by the arrangement of atoms or ions in a particular manner. A crystal is defined as a regular, long-ranged repeated arrangement of unit cells.

Crystal have a sharp melting and boiling point and they give a sharp edge on being cut with a knife.

5 0
2 years ago
Quinn accelerates her skateboard along a straight path from 0 m/s to 4.0 m/s
umka2103 [35]
  • initial velocity=u=0m/s
  • Final velocity=v=4m/s
  • Time=t=2.5s

\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \sf\longmapsto Acceleration=\dfrac{4-0}{2.5}

\\ \sf\longmapsto Acceleration=\dfrac{4}{2.5}

\\ \sf\longmapsto Acceleration=1.6m/s^2

8 0
2 years ago
When the mass is removed, the length of the cable is found to be l0 = 4.66 m. After the mass is added, the length is remeasured
zaharov [31]

Answer:

\gamma=6.07*10^5\frac{N}{m^2}

Explanation:

For a linear elastic material Young's modulus is a constant that is given by:

\gamma=\frac{F/A}{\Delta L/L_0}

Here, F is the force exerted on an object under tensio, A is the area of the cross-section perpendicular to the applied force, \Delta L is the amount by which the length of the object changes and  L_0 is the original length of the object. In this case the force is the weight of the mass:

F=mg\\F=55kg(9.8\frac{m}{s^2})\\F=539N

Replacing the given values in Young's modulus formula:

\gamma=\frac{F/\pi r^2}{(L_1-L_0)/L_0}\\\gamma=\frac{539N/\pi(0.045m)^2}{(5.31m-4.66m)/4.66m}\\\gamma=6.07*10^5\frac{N}{m^2}

6 0
2 years ago
The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How
Vlada [557]

Answer:

The value is  w =  7.54 \  m        

Explanation:

From the question we are told that

     The length of the crack is  a =  0.3 \  m

     The  frequency is  f =  30.0 \ kHz =  30 *10^{3} \  Hz

      The distance outside the cave that is being consider is  D =  100 \  m

      The speed of sound is v_s =  340 \  m/s

Generally the wavelength of the wave is mathematically represented as

        \lambda =  \frac{v}f}

=>     \lambda =  \frac{340 }{30*10^{3}}

=>     \lambda = 0.0113 \ m/s

Generally for a  single slit the path difference between the interference patterns of the sound wave and the center  is mathematically represented as  

          y =  \frac{ n *  \lambda * D}{a}

=>     y =  \frac{ 1  *  0.0113 * 100}{0.3}

=>     y = 3.77 \  m

Generally the width of the sound beam is mathematically represented as

         w =  2 *  y

=>      w =  2 *  3.77

=>      w =  7.54 \  m        

4 0
3 years ago
A charge q1 of -5.00 x 10^-9 C and a charge q2 of -2.00x 10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium po
aleksandrvk [35]

Answer:

Explanation:

Let the equilibrium position of third charge be x distance from q₁.

Force on third charge due to q₁

= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²

Force on third charge due to q₂

= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

Both the force will act in opposite direction and for balancing , they should be equal.

9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

5  / x² = 2 / ( .4 - x )²

Taking square root on both sides

2.236 / x = 1.414 / .4 - x

2.236 ( .4 - x ) = 1.414 x

.8944 - 2.236 x = 1.414 x

.8944 = 3.65 x

x = .245 m

24.5 cm

So the third charge should be at a distance of 24.5 cm from q₁ .

4 0
2 years ago
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