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strojnjashka [21]
2 years ago
13

A pickup truck is carrying a load of gravel. The driver hits a bump and gravel falls out, so that the mass if the truck is one h

alf as large after hitting the bump. If the net force on the truck does not change, how does the truck's acceleration change?
Physics
2 answers:
Serjik [45]2 years ago
7 0

Answer:

Earth's gravity would increase with the increase in Earth's mass. The driver hits a bump and the gravel falls out, so that the mass of the truck is one half as large after hitting the bump.

Explanation:

Zinaida [17]2 years ago
5 0

Answer:

Earth's gravity would increase with the increase in Earth's mass. The driver hits a bump and the gravell falls out, so that the mass of the truck is one half as large after hitting the bump.

Explanation:

hope this helps :)❤️

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FromTheMoon [43]
The wires should be attached to each screw on the light not only one
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3 years ago
5. Graph A below plots a race car's speed for 5 seconds. The car's rate of acceleration is 6 m/s^2
Georgia [21]

Answer:

The answer is below

Explanation:

We are to check if the statement is true of false. If it is false, we correct the statement.

Solution:

Acceleration is the time rate of change of velocity. It is the ratio of the change in velocity to the change in time. The acceleration can be gotten from a velocity time graph by finding the slope of the graph.

The x coordinate represent the time and the y coordinate velocity.

5) Graph A passes through the point (0, 0) and (4, 24). Therefore the acceleration (slope) is:

Acceleration = \frac{24-0}{4-0}=6\ m/s^2

This is correct.

6) Graph B is a straight line of 12 m/s. It passes through (0, 12) and (4, 12). Hence:

Acceleration = \frac{12-12}{4-0}=0\ m/s^2

This is false.

Therefore the acceleration of graph B is 0 m/s².

8 0
3 years ago
A metal rod A and a metal sphere B, on insulating stands, touch each other. They are originally neutral. A positively charged ro
ryzh [129]

Answer:

The sphere is positively charged

Explanation:

This is because when the positively charged rod is brought near the metal rod A, the electrons in metal rod A and sphere B are attracted towards it into metal rod A while the positive charges in the are repelled into sphere B. So, when the charged rod is withdrawn, and metal rod A and sphere B are separated, metal rod A is now negatively charged, but sphere B is positively charged.

So, sphere B is positively charged.

3 0
4 years ago
A rocket at rest with a mass of 942 kg is acted on by an average net force of 6,731 N upwards for 21 s. What is the final veloci
Galina-37 [17]

Answer:

Explanation:

Givens

m = 942

F = 6731

t = 21 seconds

vi = 0

vf = ?

Formula

F = m * (vf -  vi )  / t

Solution

6731 = 942*(vf - 0)/21          Multiply both sides by 21

6731 * 21 = 942*vf

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6 0
3 years ago
Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,
Marrrta [24]

Answer:

a

 Solid Wire     I  =   0.01237 \  A      

  Stranded  Wire  I_2  =   0.00978 \  A

b

  Solid Wire   R  = 0.0149 \ \Omega

   Stranded  Wire  R_1  = 0.0189 \ \Omega

Explanation:

Considering the first question

From the question we are told that

  The  radius of the first wire is  r_1  = 1.53 mm = 0.0015 \  m

  The radius of  each strand is  r_0 =  0.306 \ mm =  0.000306 \ m

  The current density in both wires is  J  =  1750 \  A/m^2

Considering the first wire

     The  cross-sectional area of the first wire is

      A   = \pi  r^2

= >  A   = 3.142 *  (0.0015)^2

= >  A   = 7.0695 *10^{-6} \  m^2

Generally the current in the first wire is    

     I  =  J*A

=>  I  =  1750*7.0695 *10^{-6}

=>  I  =   0.01237 \  A

Considering the second wire  wire

The  cross-sectional area of the second wire is

     A_1  =  19 *  \pi r^2

=>     A_1  =  19 *3.142 *  (0.000306)^2

=>  A_1  =  5.5899 *10^{-6} \  m^2

Generally the current is  

      I_2  =  J  *  A_1

=>    I_2  =   1750  *  5.5899 *10^{-6}

=>    I_2  =   0.00978 \  A

Considering question two  

 From the question we are told that

     Resistivity is  \rho  =  1.69* 10^{-8} \Omega \cdot m

     The  length of each wire  is  l =  6.25 \  m

Generally the resistance of the first wire is mathematically represented as

    R  =  \frac{\rho *  l  }{A}

=> R  =  \frac{  1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }

=> R  = 0.0149 \ \Omega

Generally the resistance of the first wire is mathematically represented as

    R_1  =  \frac{\rho *  l  }{A_1}

=> R_1  =  \frac{  1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }

=> R_1  = 0.0189 \ \Omega

3 0
3 years ago
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