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True [87]
3 years ago
13

A cyclist moves at 10 m / s for 5 minutes and at 18 m / s for 6 minutes in the same direction and direction. Get the average spe

ed of this movement.
A.14.4 m/s
B.2.5 m/s
C.25.0 m/s
D.31.6 m/s
E.None of the above
Physics
1 answer:
andre [41]3 years ago
7 0

total speed/total time=28/11=approx 2.8, just under.

makes it b, surprisingly

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Analyze how buffers allow you to eat acidic and basic foods without changing your blood pH.
blondinia [14]
Buffers neutralize the acid and the bases
7 0
3 years ago
How much work in joules is required to lift a 23 kg box up from the ground to your waist that is 1.0 meters high, carry it 6 met
PSYCHO15rus [73]

Answer:

2682

Explanation:

Work done is given by :

Work = Force x distance

         =  mg x d

So, work done in lifting the box of 23 kg up to my waist of 1 m high is :

W = mg x d

   = 23 x 9.18 x 1

   = 211.14

Now work done carrying the box horizontally 6 meters across the room is

W = mg x d

   = 23 x 9.18 x 6

   = 1266.84

Work done in placing the box on the shelf that is 5.7 m above the ground is

W = mg x d

   = 23 x 9.18 x 5.7

   = 1203.49

So the total work done is = 211.14 + 1266.84 + 1203.49

                                          = 2681.47

                                          = 2682 (rounding off)

5 0
2 years ago
What is the unit and tool to measure the volume of a plastic block?
Marina CMI [18]
Unit is m^3 or metres cubed. You need to multiply the three dimensions of the block to get the volume.
A ruler can be used to measure the edges.
8 0
3 years ago
Two identical charges q placed 2.0 mapart exert forces of magnitude 4.0 N on each other What is the value of the charge q? a) q
katen-ka-za [31]

Answer:

c) 4.2*10^{-5}C

Explanation:

Coulomb's law says that the force exerted between two charges is inversely proportional to the square of distance between them, and is given by the expression:

F=\frac{kq_{1}q_{2}}{r^{2}}

where k is a proportionality constant with the value k=9*10^{9}\frac{Nm^{2}}{C^{2}}

In this case q_{1}=q_{2}=q, so we have:

F=\frac{kq^{2}}{r^{2}}

Solving the equation for q, we have:

kq^{2}=Fr^{2}

q^{2}=\frac{Fr^{2}}{k}

q=\sqrt{\frac{Fr^{2}}{k}}

Replacing the given values:

q=\sqrt{\frac{4.0N*(2.0m)^{2}}{9*10^{-9}\frac{Nm^{2}}{C^{2}}}}

q=4.2*10^{-5}C

3 0
3 years ago
Two masses —m1 and m2— are connected by light cables to the perimeters of two cylinders of radii r1 and r2 respectively, as show
Aleksandr [31]

Answer:

Part a)

Mass of m2 is given as

m_2 = \frac{20}{3} kg

Part b)

Angular acceleration is given as

\alpha = 1.96 rad/s^2

Part c)

Tension in the rope is given as

T = 176.6 N

Explanation:

Part a)

When m1 and m2 both connected to the cylinder then the system is at rest

so we can use torque balance here

m_1g r_1 = m_2 g r_2

20 g(0.5) = m_2 g(1.5)

10 = 1.5 m_2

m_2 = \frac{20}{3} kg

Part b)

When block m_2 is removed then system becomes unstable

so force equation of mass m1

m_1g - T = m_1 a

also we have

T r_1 = I\alpha

now we have

m_1g = \frac{I a}{r_1^2} + m_1 a

a = \frac{m_1g}{\frac{I}{r_1^2} + m_1}

a = \frac{20 (9.81)}{\frac{45}{0.5^2} + 20}

a = 0.981 m/s^2

so angular acceleration is given as

\alpha = \frac{a}{r_1}

\alpha = \frac{0.981}{0.5}

\alpha = 1.96 rad/s^2

Part c)

Tension in the rope is given as

T = \frac{I\alpha}{r_1}

T = \frac{45 (1.96)}{0.5}

T = 176.6 N

7 0
3 years ago
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