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fomenos
3 years ago
6

A car is traveling at a speed of 38.0 m/s on an interstate highway where the speed limit is 75.0 mi/h. Is the driver exceeding t

he speed limit?
Physics
1 answer:
borishaifa [10]3 years ago
4 0

Answer:

yes driver exceed the car speed

Explanation:

given data

speed of car = 38 m/s

speed limit = 75.0 mi/h

to find out

Is the driver exceeding the speed limit

solution

we know car speed is 38 m/s and limit is 75 mi/hr

so for compare the speed limit we convert limit and make them same

as we know

1 m/s = 2.236 mi/hr

so

car speed 38 m/s = 38 × 2.236 = 85.003 mi/hr

as this car speed is exceed the speed limit that is 75 mi/hr

yes driver exceed the car speed

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PLEASEE I NEED HELP FAST!!! .Study the scenario.A small container of water with a low temperature is poured into a large contain
RSB [31]

Answer:

that best describes the process is C

Explanation:

This problem is a calorimeter process where the heat given off by one body is equal to the heat absorbed by the other.

Heat absorbed by the smallest container

             Q_c = m ce (T_{f}-T₀)

Heat released by the largest container is

              Q_a = M ce (T_{i}-T_{f})

how

        Q_c = Q_a

       m (T_{f}-T₀) = M (T_{i} - T_{f})

Therefore, we see that the smaller container has less thermal energy and when placed in contact with the larger one, it absorbs part of the heat from it until the thermal energy of the two containers is the same.

Of the final statements, the one that best describes the process is C

since it talks about the thermal energy and the heat that is transferred in the process

8 0
4 years ago
Two dump trucks each have a mass of 1,500 kg. The distance of the dump truck
xxTIMURxx [149]

Answer:

6.00 x 10⁻⁸N

Explanation:

Given parameters:

Mass of each dump trucks  = 1500kg

Distance between them  = 50m

Unknown:

New gravitational force between them = ?

Solution:

From Newton's law of universal gravitation,

        F = \frac{G m1 m2}{r^{2} }  

F is the gravitational force

G is the universal gravitation constant

m is the mass

r is the distance

           F  = \frac{6.67 x 10^{-11} x 1500  x 1500}{50^{2} }    = 6.00 x 10⁻⁸N

4 0
3 years ago
A positive charge is moved from point A to point B along an equipotential surface. How much work is performed or required in mov
Nitella [24]

Answer:

No work is performed or required in moving the positive charge from point A to point B.

Explanation:

Lets take

Q= Positive charge which move from  point A to point B along

Voltage difference,ΔV =V₁ - V₂  

The work done

W = Q . ΔV

Given that  charge is moved from point A to point B along an equipotential surface.It means that voltage  difference is zero.

ΔV = 0

So

W = Q . ΔV

W = Q x 0

W= 0 J

So work is zero.

5 0
3 years ago
In a ballistics test, a 52g bullet hits a sand bag and stops after moving 1.34 m. If the initial bullat
olya-2409 [2.1K]

Answer:

Friction force on the bullet is 58.7 N opposite to its velocity

Explanation:

As we know that initial speed of the bullet is 55 m/s

after travelling into the sand bag by distance d = 1.34 m it comes to rest

so final speed

v_f = 0

now we can use kinematics top find the acceleration of the bullet

v_f^2 - v_i^2 = 2 a d

so we have

0 - 55^2 = 2(a)(1.34)

a = -1128.7 m/s^2

now by Newton's II law we know that

F = ma

so we have

F = (0.052)(-1128.7)

F = -58.7 N

8 0
3 years ago
Un avión de rescate en Alaska deja caer un paquete de provisiones a un grupo de exploradores extraviados. Si el avión viaja hori
posledela

Answer:

180.4 m

Explanation:

The package in relation to the point where it was released falls a certain distance that is calculated by applying the horizontal motion formulas , as the horizontal speed of the plane and the height above the ground are known, the time that It takes the package to reach its destination and then the horizontal distance (x) is calculated from where it was dropped, as follows:    

$V_{ox}=v_x = 40 \ m/s$

   h = 100 m  

    x =?

     Height formula h:

     $h=g \times \frac{t^2}{2}$

      Time t is cleared:

     $t = \sqrt{\frac{2h}{g}}$

      $t = \sqrt{\frac{2 \times 100}{9.8}}$

      t = 4.51 sec

 Horizontal distance formula x:

       $x=V_x \times t$

        x = 40 m / sec x 4.51 sec

        x = 180.4 m

4 0
3 years ago
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