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Svetach [21]
2 years ago
6

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23

.5 m/s is h = 2 + 23.5t − 4.9t2 after t seconds. (Round your answers to two decimal places.) (a) Find the velocity after 2 s and after 4 s. v(2) = m/s v(4) = m/s
Physics
1 answer:
Burka [1]2 years ago
8 0

Answer:

a) v(2) = 3.9\,\frac{m}{s}, b) v(4) = -15.7\,\frac{m}{s}

Explanation:

a) The equation for vertical velocity is obtained by deriving the function with respect to time:

v(t) = 23.5 -9.8\cdot t

The velocities at given instants are, respectivelly:

v(2) = 3.9\,\frac{m}{s}

v(4) = -15.7\,\frac{m}{s}

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A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w
Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, a=3\ m/s^2

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.  

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

m=\dfrac{W}{g}

m=\dfrac{150}{9.8}

m=15.3\ kg

Frictional force is given by :

F=ma

F=15.3\times 3

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

8 0
2 years ago
The velocity of a car increases from 2.0 m/s to 16.0 m/s in a time period of 3.5 s. What was the average acceleration?
dangina [55]

Answer:

the acceleration is a=3ms^-2

Explanation:

8 0
1 year ago
Read 2 more answers
A metal, M, forms an oxide having the formula M2O3 containing 52.92% metal by mass. Determine the atomic weight in g/mole of the
irina [24]

Answer:

The atomic weight in g/mole of the metal (molar mass) is 8.87.

Explanation:

To begin, it is possible to assume that, as a sample, it has 100 g of the compound. This means that:

  • 52.92% metal: 52.92 g M
  • 47.80% oxygen: 47.80 g O

 Using the molar mass of oxygen, which is 16 g / mol, it is possible to calculate the amount of moles of oxygen present in the sample using the rule of three:

moles of oxygen=\frac{47.8g*1mol}{16g}

moles of oxygen=2.9875

The chemical formula of metal oxide tells you that:

2 M⁺³ + 3 O²⁻ ⇒ M₂O₃

In the previous equation you can see that you need 3  oxygen anions to react with two metal cations. Then:

2.9875 moles of oxygen*\frac{2 moles of metal M}{1 mol of oxygen} = 5.975 moles of metal M

You have  52.92 g  of metal in the sample, then the molar mass of the metal is:

molar mass=\frac{52.92 g}{5.975 mol}

molar mass≅ 8.87 g/mol

<u><em> The atomic weight in g/mole of the metal (molar mass) is 8.87.</em></u>

The closest match to this value is Beryllium (Be), which has an atomic mass of 9.0122 g / mol.

3 0
2 years ago
An automobile traveling at 25.0 km/h along a straight, level road accelerates to 65.0 km/h in 6.00 s. what is the magnitude of t
USPshnik [31]
Note that
1 km/h = (1000 m)/(3600 s) = 0.27778 m/s

Initial velocity, v₁ = 25 km/h = 6.9444 m/s
Final velocity, v₂ = 65 km/h = 18.0556 m/s

Time interval, dt = 6 s.

Calculate average acceleration.
a = (v₂ - v₁)/dt
   = (18.0556 - 6.9444 m/s)/(6 s)
   = 1.852 m/s²

Answer:
The average acceleration is 1.85 m/s² (nearest hundredth)
3 0
2 years ago
Suppose you are talking by interplanetary telephone to your friend, who lives on the Moon. He tells you that he has just won a n
Savatey [412]

Answer:

The friend on moon will be richer.

Explanation:

We must calculate the mass of gold won by each person, to tell who is richer. For that purpose we will use the following formula:

W = mg

m = W/g

where,

m = mass of gold

W = weight of gold

g = acceleration due to gravity on that planet

<u>FOR FRIEND ON MOON</u>:

W = 1 N

g = 1.625 m/s²

Therefore,

m = (1 N)/(1.625 m/s²)

m(moon) = 0.6 kg

<u>FOR ME ON EARTH</u>:

W = 1 N

g = 9.8 m/s²

Therefore,

m = (1 N)/(9.8 m/s²)

m(earth) = 0.1 kg

Since, the mass of gold on moon is greater than the mass of moon on earth.

<u>Therefore, the friend on moon will be richer.</u>

7 0
2 years ago
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