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vichka [17]
2 years ago
15

Tyson throws a shot put ball weighing 7.26 kg. At a height of 2.1 m above the ground, the mechanical energy of the ball is 172.1

J. What is the velocity of the ball at the given point? m/s
Physics
2 answers:
max2010maxim [7]2 years ago
7 0

Answer:

2.5 m/s

Explanation:

Mechanical energy is the sum of the potential and kinetic energy.

E = PE + KE

E = mgh + ½mv²

172.1 J = (7.26 kg) (9.8 m/s²) (2.1 m) + ½ (7.26 kg) v²

v = 2.5 m/s

Mumz [18]2 years ago
6 0

Answer:

2.5 m/s

Explanation: i jus did this on edg. its right:)

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How much heat does it take to raise the temperature of 7.0 kg of water from
aksik [14]

Answer:

non of the above

Explanation:

Quantity of heat = mass× specific heat× change in temperature

m= 7kg c= 4.18 temp= 46-25=21°

.......H= 7×4.18×21= 614.46kJ

6 0
2 years ago
Read 2 more answers
A 0.50-kg block slides across a tabletop with an initial velocity of 20 cm/s and comes to rest in a distance of 70 cm. Find the
mars1129 [50]

Answer:

FAE= 0.014 N

Explanation:

The KE of block is decreased because of the slowing action of the friction force .

Change in KE of block = work done on block by friction ƒ

⠀ ➪ ½mu²ƒ - ½mu²i = Fƒs cos θ

Because the friction force on the block is opposite in direction to the displacement , cos θ = -1

➢ Using Uƒ = 0 , Vƒ = 0.20 m/s , and s = 0.70 m

✒ We find ,

➪½mu²ƒ - ½mu²i = Fƒs cos θ

➪0-½ (0.50 kg) (0.20 m/s)² = (Fƒ) (0.70 m) (-1)

➪ Fƒ = 0.014 N

Hope this helped, can i pls have brainliest

6 0
2 years ago
Read 2 more answers
An empty glass beaker has a mass of 103 g. When filled with water, it has a total mass of 361g.
aivan3 [116]

Answer:

0.96 gcm¯³

Explanation:

From the question given above, the following data were obtained:

Mass of empty beaker = 103 g

Mass of beaker + water = 361 g

Mass of beaker + oil = 351 g

Density of water = 1 gcm¯³

Density of cooking oil =?

Next, we shall determine the mass of water. This can be obtained as follow:

Mass of empty beaker = 103 g

Mass of beaker + water = 361 g

Mass of water =?

Mass of water = (Mass of beaker + water) – (Mass of empty beaker)

Mass of water = 361 – 103

Mass of water = 258 g

Next, we shall determine the volume of the beaker. This can be obtained by calculating the volume of water in the beaker.

Density of water = 1 gcm¯³

Mass of water = 258 g

Volume of water =?

Density = mass /volume

1 = 258 / volume

Cross multiply

1 × volume = 258

Volume of water = 258 cm³

Thus the volume of the beaker is 258 cm³.

Next, we shall determine the mass of the cooking oil. This can be obtained as follow:

Mass of empty beaker = 103 g

Mass of beaker + oil = 351 g

Mass of cooking oil =?

Mass of cooking oil = (Mass of beaker + oil) – (Mass of empty beaker)

Mass of cooking oil = 351 – 103

Mass of cooking oil = 248 g

Finally, we shall determine the density of the cooking oil. This can be obtained as follow:

Mass of cooking oil = 248 g

Volume of the beaker = 258 cm³

Density of cooking oil =?

Density = mass / volume

Density = 248 / 258

Density of cooking oil = 0.96 gcm¯³

7 0
2 years ago
A 0.46-kg cord is stretched between two supports, 7.2 m apart. When one support is struck by a hammer, a transverse wave travels
Kryger [21]

Answer:

      T = 6.0 N

Explanation:

given,

mass of the cord = 0.46 Kg

length of the supports = 7.2 m

time taken to travel = 0.74 s

tension in the chord = ?

using formula for tension calculation

T = \dfrac{v^2.m}{l}

v = \dfrac{l}{s}

v = \dfrac{7.2}{0.74}

v = 9.73 m/s

now, calculation of tension

T = \dfrac{9.73^2\times 0.46}{7.2}

      T = 6.0 N

The tension in the cord is equal to 6.0 N.

7 0
2 years ago
A man walking on a tightrope carries a long a pole which has heavy items attached to the two ends. If he were to walk the tight-
katen-ka-za [31]

Answer:

 I_weight = M L²

this value is much larger and with it it is easier to restore balance.I

Explanation:

When man walks a tightrope, he carries a linear velocity, this velocity is related to the angular velocity by

            v = w r

For man to maintain equilibrium needs the total moment to be zero

             ∑τ = I α

              S  τ = 0

The forces on the home are the weight of the masses, the weight of the man and the support on the rope, the latter two are zero taque the distance to the center of rotation is zero.

Therefore the moment of the masses and the open is the one that must be zero.

If the man carries only the bar, we could approximate it by two open one on each side of the axis of rotation formed by the free of the rope

              I = ⅓ m L² / 4

As the length of half the length of the bar and the mass of the bar is small, this moment is small, therefore at the moment if there is some imbalance it is difficult to recover.

If, in addition to the opening, each of them carries a specific weight, the moment of inertia of this weight is

             I_weight = M L²

this value is much larger and with it it is easier to restore balance.

5 0
2 years ago
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