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Sonja [21]
4 years ago
13

Which statement about the horizontal distance covered by a projectile launched at an angle less than 90° is true?

Physics
1 answer:
Step2247 [10]4 years ago
8 0
The answer is A because the distance covered is equal time intervals is the same or equal.
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Determine the thrust that a boat with a volume of 1.2m³ receives when it is stranded at sea. The density of seawater is 1020kg /
puteri [66]

Answer:

The maximum possible up-thrust on the boat is 11,995.2 N

Explanation:

According to Archimedes' principle, the thrust received by an object immersed a fluid is equal to the weight of the fluid displaced;

The given parameter of the boat in sea water are;

The volume of the boat = 1.2 m³

The density of seawater = 1020 kg/m³

Density = Mass/Volume

Therefore, Mass = Density × Volume

The maximum volume of water that the boat displaces = 1.2 m³

The mass of the water displaced by the boat = (Density of seawater) × (Volume of seawater displaced)

∴ The maximum possible mass of the water displaced by the boat = 1.2 m³ × 1020 kg/m³ = 1224 kg

The maximum possible mass of the water displaced by the boat, m = 1224 kg

Weight = Mass, m × g

Where;

g = The acceleration due to gravity = 9.8 m/s²

The up-thrust on the boat = The weight of the seawater displaced

∴ The maximum possible up-thrust on the boat = m × g = 1224 kg × 9.8 m/s² = 11,995.2 N

The maximum possible up-thrust on the boat = 11,995.2 N.

3 0
3 years ago
A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m
madam [21]

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR   where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂  = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

4 0
3 years ago
Betty weighs 400 N and she is sitting on a playground swing seat that hangs 0.21 m above the ground. Tom pulls the swing back an
disa [49]

Answer:

4.15 m/s

Explanation:

As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:

ΔE = ΔK + ΔU =0

If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:

h₀ = 1.09 m -0.21 m = 0.88 m

⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J

As Uf = 0, ΔU = Uf -U₀ = -352 J

As the swing starts from rest, K₀=0, so we can say:

ΔK = Kf = \frac{1}{2} *m*vf^{2}  (1)

As ΔK = -ΔU ⇒ ΔK = 352 J (2)

From (1) and (2) we can solve for vf, as follows:

vf = \sqrt{\frac{2*352J}{40.8kg}} = 4.15 m/s

So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.

5 0
3 years ago
Divers in Acapulco dive from a cliff that is 65 m high. If the rocks
Ghella [55]

Answer:

did you have the same answer to get the best

6 0
2 years ago
An electron is brought from rest infinitely far away to rest at point P located at a distance of 0.041 m from a fixed charge q.
torisob [31]

Answer:

The electric potential in volts is 1.618 x 10⁻¹⁷ V

Explanation:

The electric potential, in volts, at point P, can be calculated as follow;

Electric potential is the work done in moving a unit positive charge from infinity to a particular point in the electric field.

Thus, the work done in this process in moving the charge to point p is 101eV.

Convert this Volts = 101 × 1.602 x 10⁻¹⁹ V

                              = 1.618 x 10⁻¹⁷ V

Therefore, the electric potential in volts is 1.618 x 10⁻¹⁷ V

3 0
3 years ago
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