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larisa86 [58]
1 year ago
11

A battleship simultaneously fires two shells toward two identical enemy ships. One shell hits ship A, which is close by, and the

other hits ship B, which is farther away. The two shells are fired at the same speed. Assume that air resistance is negligible and that the magnitude of the acceleration due to gravity is .Now, consider both shells are fired at an angle greater than 45 degrees with respect to the horizontal. Remember that enemy ship A is closer than enemy ship BWhich shell is launched with a greater horizontal velocity, ?a. Ab. Bc. Both shells are launched with the same horizontal velocity.
Physics
1 answer:
luda_lava [24]1 year ago
8 0

Answer:

both cannonballs hit the ships with the same horizontal speed

Explanation:

Hello!

A parabolic motion is characterized in that its vertical component in Y is constantly changing, this is due to the constant downward acceleration of gravity.

When the movement starts the speed at Y is maximum, then when it reaches its maximum height point its speed is zero, and finally it begins to increase downwards until it touches the floor.

On the other hand, the horizontal speed remains constant AS THERE IS NO ACCELERATION IN HORIZONTAL DIRECTION.

therefore both cannonballs hit the ships with the same horizontal speed

regards!

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A rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two sides of the loop are
Dahasolnce [82]

Answer:

(a) 0.05 Am^2

(b) 1.85 x 10^-3 Nm

Explanation:

width, w = 10 cm = 0.1 m

length, l = 20 cm = 0.2 m

Current, i = 2.5 A

Magnetic field, B = 0.037 T

(A) Magnetic moment, M = i x A

Where, A be the area of loop

M = 2.5 x 0.1 x 0.2 = 0.05 Am^2

(B) Torque, τ = M x B x Sin 90

τ = 0.05 x 0.037 x 1

τ = 1.85 x 10^-3 Nm

4 0
2 years ago
1. A2 .7-kg copper block is given an initial speed of 4.0m/s on a rough horizontal surface. Because of friction, the block final
tatyana61 [14]

Answer:

A. Increase in temperature is 0.0176 degree Celsius. b. the remaining energy will be lost.

Explanation:

The mass of copper block = 7kg

Initial speed = 4.0 m/s

Specific heat of copper = 0.385 j/g degree Celcius.

a. The increase in temperature is calculated below:

\text{Change in kinetic energy} = \frac{1}{2}mv^{2} \\=  \frac{1}{2} \times 2.7 \times 4^{2} = 21.6 J \\

85% of energy is converted into internal energy.

mc\DeltaT = 21.6 \times 0.85 \\2.7 \times 385 \times \Delta T = 21.6 \times 0.85 \\\Delta T = 0.0176 degree \ celsius

b. The remaining  15 per cent of kinetic energy will be lost and it will be changed into other forms.

7 0
1 year ago
A 60-μC charge is held fixed at the origin and a −20-μC charge is held fixed on the x axis at a point x = 1.0 m. If a 10-μC char
Aleksandr [31]

Answer:

Ek =  8,79 [J]

Explanation:

We are going to solve this problem, using  the energy conservation principle

State 1 or initial state (charges at rest t=0)

E₁  = Ek  + U₁

As charge are at rest Ek = 0

And  U₁ has two components

U₁₂   = K * Q₁*Q₂ / 0,4          and    U₃₂  = K*Q₃*Q₂ / 0,6

U₁₂  = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,4  ⇒ U₁₂ = 9*60*10*10⁻³/0,4

U₃₂ =  - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,6  ⇒ U₃₂ = - 9*20*10*10⁻³/0,6

U₁₂ = 540*10⁻2/0,4 [J]   ⇒13,5 [J]

U₃₂ = - 180*10⁻² /0,6 [J] ⇒ - 3 [J]

Then   E₁ = E₁₂ + E₃₂    

E₁ = 10,5 [J]

At  the moment of Q₂ passing x = 40 cm  or 0,4 m

E₂ = Ek + U₂

We can calculate the components of U₂ in this new configuration

U₂  =  U₁₂  + U₃₂

U₁₂  = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,7   ⇒  U₁₂ = 9*60*10*10⁻³/0,7

U₁₂ = 540*10⁻²/0,7       U₁₂ = 7,71 [J]

U₃₂ =  - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,3  ⇒ U₃₂ = -  9*20*10*10⁻³/0,3

U₃₂ = -  9*20*10⁻²/0,3  

U₃₂ = - 6

U₂ = 7,71 -6

U₂ = 1,71 [J]

Then as  

E₂  = Ek + U₂  and  E₂ = E₁

Then

Ek + U₂ = E₁

Ek =  10,5 - U₂    

Ek  = 10,5 - 1,71

Ek =  8,79 [J]

5 0
1 year ago
"A block of metal weighs 40 N in air and 30 N in water. What is the buoyant force on the block due to the water? The density of
Alja [10]

Answer:

buoyant force on the block due to the water= 10 N

Explanation:

We know that

buoyant force(F_B) on a block= weight of the block in air (actual weight) - weight of block in water.

Given:

A block of metal weighs 40 N in air and 30 N in water.

F_B =  40-30= 10 N

therefore,  buoyant force on the block due to the water= 10 N

6 0
1 year ago
Read 2 more answers
A goat enclosure is in the shape of a right triangle. One leg of the enclosure is built against the side of the barn. The other
san4es73 [151]

Answer:

16,18,22

Or

1,3,7

Explanation:

The detailed explanation is contained in the image attached. The lengths are found using Pythagoras theorem and the two lengths reflects the two values of x yielded by the quadratic equation

8 0
2 years ago
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