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Molodets [167]
2 years ago
7

To test the resiliency of its bumper during low-speed collisions, a 3 690-kg automobile is driven into a brick wall. The car's b

umper behaves like a spring with a force constant 8.00 106 N/m and compresses 3.86 cm as the car is brought to rest. What was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall?
Physics
1 answer:
Mariana [72]2 years ago
8 0

Answer:

1.8 m/s

Explanation:

m = mass of the automobile = 3690 kg

k = force constant of the spring = 8 x 10⁶ N/m

x = compression of the spring = 3.86 cm = 0.0386 m

v = speed of the car before impact

Using conservation of energy

Kinetic energy of automobile before impact = Spring potential energy of bumpers

(0.5) m v² = (0.5) k x²

(3690) v² = (8 x 10⁶) (0.0386)²

v = 1.8 m/s

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A car is initially driving at 30 m/s. It hits a large pothole, after which it is traveling in the same direction but at 25 m/s.
Bogdan [553]

Answer:

The time of Mars is 1.65 times larger on Mars than on Earth

Explanation:

The equation that describes the system is the final speed is equal to the speed minos the speed lost by the collision with the porhole

       Vf = Vo - V pothole

B) let's transform the weight of free groin system and N international system

      1 N = 0.2248 lb

      2.8 lbs (1N / 0.2248lbs) = 12.5 N

c) Kinematic equations are the same in all inertial systems, Mars and Earth, so we can use the height equation, with zero initial velocity

                   

        Y = Vo t - ½ g t²

        Y = - ½ g t²

        t = √ 2Y / g

     

Mars

         gm = 0.37g

         gm = 0.37 9.8

         gm = 3,626 m / s²

         t = √( 2 1.9 / 3.626 )

         t = 1.02 s

Earth

         t = √( 2 1.9 / 9.8)

         t = 0.62 s

To make the comparison of time we are the relationship between the two

         tm / te = 1.02 / 0.62

         tm / te = 1.65

The time of Mars is 1.65 times larger on Mars than on Earth

7 0
2 years ago
Read 2 more answers
Part b suppose the magnitude of the gravitational force between two spherical objects is 2000 n when they are 100 km apart. what
kobusy [5.1K]
<span>b) The force with a distance of 150 km is 889 N c) The force with a distance of 50 km is 8000 N This question looks like a mixture of a question and a critique of a previous answer. I'll attempt to address the original question. Since the radius of the spherical objects isn't mentioned anywhere, I will assume that the distance from the center of each spherical object is what's being given. The gravitational force between two masses is given as F = (G M1 M2)/r^2 where F = Force G = gravitational constant M1 = Mass 1 M2 = Mass 2 r = distance between center of masses for the two masses. So with a r value of 100 km, we have a force of 2000 Newtons. If we change the distance to 150 km, that increases the distance by a factor of 1.5 and since the force varies with the inverse square, we get the original force divided by 2.25. And 2000 / 2.25 = 888.88888.... when rounded to 3 digits gives us 889. Looking at what looks like an answer of 890 in the question is explainable as someone rounding incorrectly to 2 significant digits. If the distance is changed to 50 km from the original 100 km, then you have half the distance (50/100 = 0.5) and the squaring will give you a new divisor of 0.25, and 2000 / 0.25 = 8000. So the force increases to 8000 Newtons.</span>
8 0
2 years ago
Read 2 more answers
A dart is inserted into a spring-loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is c
bezimeni [28]

Answer:

The second dart leaves the gun two times as faster than the first one.

Explanation:

Assuming no energy loss during the spring-dart energy transfer, we have by the conservation of energy principle

U_s = K_d \\ \frac{1}{2} kx^2 = \frac{1}{2}mv^2 \\ v = \sqrt{\frac{k}{m}x^2}.

Given an arbitrary x and its double, 2x, launch velocities are

v_1 = \sqrt{\frac{k}{m}x^2} \text{ and} \\ v_2 = \sqrt{\frac{k}{m}\left(2x\right)^2} = \sqrt{\frac{k}{m}4x^2} = 2\sqrt{\frac{k}{m}x^2} = \mathbf{2v_1}.

7 0
3 years ago
Calculate the equivalent of 20 degrees Celsius in degrees Fahrenheit and Kelvin.
alekssr [168]

Answer:

68 °F, 293.15 K

Explanation:

Fahrenheit, Kelvin and Celsius are the different scales of temperature in which temperature is measured.

Given : T = 20°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

<u>T = (20 + 273.15) K = 293.15 K </u>

The conversion of T( °C) to T(F) is shown below:

T (°F) = (T (°C) × 9/5) + 32

So,

<u>T (°F) = (20 × 9/5) + 32 = 68 °F</u>

3 0
3 years ago
3.1
Ilia_Sergeevich [38]

Answer:

The value is  C = 30729\  c

Explanation:

From the question we are told that

  The power rating of the stove is  P = 4.4 KW  = 4.4 *10^{3} \  W

   The duration of its use everyday is t_1  = 70 \ minutes  = 1.167 \ hours    

    The rating of the light bulbs is P_2'  = 150 W

   The number is n = 7

   The power rating of the total  bulb is  P_2 = 7 * 150 = 1050 \  W

    The duration of its use everyday is  t_2  = 7 hours

    The power rating of miscellaneous appliance P_3 = 1.8 \ KW = 1.8 *10^{3} \  W

    The duration of its use everyday is t_3 = 1 hour

     The power rating of hot water P = 4 KW  = 4 *10^{3} \  W

      The duration of its use everyday is t_4 = 120 \ minutes  = 2 hours

Generally the total electrical energy used in 1 month is mathematically represented as

    E = P_1 * t_1 * 30 + P_2 * t_2 * 30 + P_3 * t_3 * 30+ P_4 * t_4 * 30

=>  E = 4.4*10^{3} * 1.167  * 30 + 1050 * 7 * 30 + 1.8 *1 * 30+ 4*10^{3} * 2 * 30

=>  E = 614598  \ W \cdot h

=>   E = 614.6  \  K W \cdot h

Generally the monthly electricity  bill is mathematically represented as

    C = 614.6 * 50

=> C = 30729\  c

8 0
2 years ago
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