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liraira [26]
3 years ago
15

Which graph illustrates constant speed and velocity?

Physics
2 answers:
boyakko [2]3 years ago
8 0

The correct graph is <u>D</u>.

The graph <em>A</em> is a straight line sloping downwards and it shows that the speed of the body is decreasing at a constant rate. Therefore, this s a graph of a body that is under a constant deceleration.

The graph B is a straight line which slopes upwards. Hence the graph shows that the speed of the body increases at a constant rate. Therefore, this is a graph of a body that is accelerating at a constant rate.

The graph C is curved line, which curves upwards. The slope of the curve increases with time. This is therefore, a graph of a body which is under increasing acceleration.

The graph D, however is a straight line parallel to the time axis. The speed of the body has the same value at all times. Therefore, Graph D is the graph which shows the motion of a body with constant speed.

denpristay [2]3 years ago
5 0

Answer choice D is correct.

When you are thinking about a "constant" on a graph, it usually means that is is a straight line that is either horizontal or vertical. In this case the only graph that as both a constant speed and velocity is D which is why it is correct.

A has a negative constant acceleration

B has a positive acceleration

C is not constant, but is moving at a positive acceleration.

Hope this helps some!

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A 0.76-kg block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the first one
sergey [27]

Answer:

1.54 kg

Explanation:

mass of first block (m) = 0.76 kg

acceleration due to gravity (g) = 9.8 m/s

what is the mass (m) of the second block

  • force = kx ⇒ mg = kx

        mg = kx

        where m is the mass, g is the acceleration due to gravity, k is the  

        spring constant and x is the extension

        0.76 x 9.8 = kx

       7.5 = kx

        k = 7.5/x ... equation 1

  • when a second block is attached to the first one the amount of stretch triples (this means that extension (x) = 3x)

        therefore the new mass becomes m + 0.76 and the extension  

        becomes 3x

        with the new mass and extension, mg = kx now becomes

        (m+0.76)g = k(3x) ... equation 2

        Recall that k = 7.5/x from equation 1, substituting this value of k into      

        equation 2 we have

         (m+0.76)g =  \frac{7.5}{x} × (3x)

         (m+0.76)g =  7.5 × 3

          substituting the value of g = 9.8 m/s^{2}

         (m + 0.76) x 9.8 = 7.5 x 3

          m + 0.76 = 22.5 ÷ 9.8

          m + 0.76 = 2.3

          m = 2.3 - 0.76 = 1.54 kg

       

         

       

4 0
3 years ago
Two point charges are 10.0cm apart and have charges of 2.0uC and -2.0uC, respectively. What is the magnitude of the electric fie
elena-14-01-66 [18.8K]
The electric field generated by a point charge is given by:
E= k_e \frac{Q}{r^2}
where
k_e = 8.99 \cdot 10^9 Nm^2 C^{-2} is the Coulomb's constant
Q is the charge
r is the distance from the charge

We want to know the net electric field at the midpoint between the two charges, so at a distance of r=5.0 cm=0.05 m from each of them. 

Let's calculate first the electric field generated by the positive charge at that point:
E_1=k_e  \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =+7.19 \cdot 10^6 N/C
where the positive sign means its direction is away from the charge.

while the electric field generated by the negative charge is:
E_2=k_e \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(-2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =-7.19 \cdot 10^6 N/C
where the negative sign means its direction is toward the charge.

If we assume that the positive charge is on the left and the negative charge is on the right, we see that E1 is directed to the right, and E2 is directed to the right as well. This means that the net electric field at the midpoint between the two charges is just the sum of the two fields:
E_{tot} =E_1 + E_2 = 7.19 \cdot 10^6 N/C+7.19 \cdot 10^6 N/C=1.44 \cdot 10^7 N/C
3 0
3 years ago
A 1300 kg car traveling with a speed of 3.5 m/s executes a turn with a 8.5 m radius of curvature.
Y_Kistochka [10]

Answer:

1.4 m/s/s (2.s.f)

Explanation:

The formula for centripetal acceleration is:

a=\frac{v^{2} }{r}, where v is velocity and r is the radius.

In the question we are given the information that the car has a mass of 1300kg, a velocity of 2.5m/s, and a turn radius of 8.5m which are all the values we need. Therefore we can simply substitute in the values to solve the question:

a=\frac{3.5^{2} }{8.5} \\a=1.4

Therefore the centripetal acceleration of the car is 1.4m/s/s. (2.s.f)

Hope this helped!

7 0
3 years ago
What is the graph of the relationship between the volume of a gas at a constant pressure
Bogdan [553]
We know that P1V1 = P2V2, if there is a constant pressure, then the P1 and P2 can cancel out, so it is V1=V2 that is whats left.
6 0
3 years ago
A temperature of 50F is equal to c
Hitman42 [59]

10.00 °C this is the right answer need more question feel free to post

5 0
3 years ago
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