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2 years ago

What is the mass of a bicycle that has 90 kg.m/s of momentum at a speed of 6 m/s?

Physics

1 answer:

OLEGan [10]2 years ago

6 0

Answer:

m = 15 kg

Explanation:

p = m × v

90 = m × 6

6m = 90

m = 90/6

m = 15 kg

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Write a note on unity of ant

Tom [10]

Answer: When a pathogen enters their colony, ants change their behavior to avoid the outbreak of disease. In this way, they protect the queen, brood and young workers from becoming ill. These results, from a study carried out in collaboration between the groups of Sylvia Cremer at the Institute of Science and Technology Austria (IST Austria) and of Laurent Keller at the University of Lausanne, are published today in the journal Science.

Explanation: search for it.

3 0

2 years ago

A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s

Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8 $^{-3}$

Explanation:

$m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg$

$v_{bullet}=341\frac{m}{s}$

Momentum of the motion the first part of the motion have a momentum that is:

$P_{1}=m_{bullet}*v_{bullet}$

$P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529$

The final momentum is the motion before the action so:

a).

$P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}$

$P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}$

$P_{1}=P_{2}$

$2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}$

b).

kinetic energy

$K=\frac{1}{2}*m*(v)^{2}$

Kinetic energy after

$Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J$

Kinetic energy before

$Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J$

Ratio = $\frac{Ka}{Kb}$

$R=\frac{1.14}{406.4}\\R=2.8x10^{-3}$

3 0

3 years ago

Suppose a 4.0-kg projectile is launched vertically with a speed of 8.0 m/s. What is the maximum height the projectile reaches?

eduard

Answer:

h = 3.3 m (Look at the explanation below, please)

Explanation:

This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy

Kinetic energy = $\frac{1}{2}$ m $v^{2}$

Plug in the numbers = $\frac{1}{2}$ (4.0)( $8^{2}$ )

Solve = 2(64) = 128 J

Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.

Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)

Kinetic energy = Potential Energy

128 J = 39.2h

h = 3.26 m

h= 3.3 m (because of significant figures)

7 0

2 years ago

Aiden takes 0.10s to throw a baseball, which leaves his hand with a velocity of 49m/s. The balls acceleration is

Brums [2.3K]

Since the ball was not moving before it let Aiden's hand, the formula used to calculate the acceleration is
$a = \frac{v}{t}$
, where a is acceleration, v is velocity and t is the time. We put them in the formula and get
$a = \frac{49}{0.1} \\ a = \frac{490}{1} \\ a = 490$
The acceleration is 490 m/s^2

7 0

3 years ago

When performing the spike the ball should be hit with the ?

VMariaS [17]

Answer:

When you are performing spike it's most effective to strike the ball from the right or left side at a sharp downward angle. Whether you are spiking the ball from the right or left front position, position yourself behind the 10-foot line (attack line), which is the line that is about four steps away from the net.

5 0

3 years ago