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vazorg [7]
3 years ago
7

A cannonball is fired horizontally at the same time a ball being dropped from the same height How do the times it takes them to

hit the ground compare?
Physics
1 answer:
ipn [44]3 years ago
3 0

Answer:

The cannonball and the ball will both take the same amount of time before they hit the ground.

Explanation:

For a ball fired horizontally from a given height, there is only a vertical acceleration on it towards the ground. This acceleration is equal to the acceleration due to gravity (g = 9.81 m/s^2). A ball dropped from a height will also only experience the same vertical acceleration downwards which is also equal to g = 9.81 m/s^2.

Therefore both the cannonball and the ball will take the same amount of time to hit the ground if they are released/fired from the same height.

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Explain how do winds cause surface currents?
igor_vitrenko [27]

Answer:

As wind or an ocean current moves, the Earth spins underneath it. ... The Coriolis effect bends the direction of surface currents to the right in the Northern Hemisphere and left in the Southern Hemisphere.

Explanation:

The Coriolis effect causes winds and currents to form circular patterns.

8 0
3 years ago
Accelerates uniformly at 2.0 ms2 for 10.0s. Calculate its final velocity​
Crazy boy [7]

Answer:

The distance is

=

7

m

Explanation:

Apply the equation of motion

s

(

t

)

=

u

t

+

1

2

a

t

2

The initial velocity is

u

=

0

m

s

−

1

The acceleration is

a

=

2

m

s

−

2

Therefore, when

t

=

3

s

, we get

s

(

3

)

=

0

+

1

2

⋅

2

⋅

3

2

=

9

m

and when

t

=

4

s

s

(

4

)

=

0

+

1

2

⋅

2

⋅

4

2

=

16

m

Therefore,

The distance travelled in the fourth second is

d

=

s

(

4

)

−

s

(

3

)

=

16

−

9

=

7

m

4 0
2 years ago
According to the Law of the Conservation of Matter, if you dissolve 25 grams of sugar into 150 grams of water, the mixture shoul
levacccp [35]
19grams becauce if u
7 0
2 years ago
How do I calculate the tension in the horizontal string?
matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

T_1\cos 30\degree-m\cdot g=0

Solve for T₁,

T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N

Now, we use the second equation to find the tension in the horizontal string,

T_2-T_1\sin 30\degree=0

Solve for T₂,

T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

8 0
1 year ago
What is the speed of a horse in meters per second that runs a distance of 1.2 miles in 2.4 minutes​
Ilia_Sergeevich [38]
Time t=2.4 minutes=2.4×60=144 seconds
distance s=1.2 miles=1.2×1609=1930.8 meters
speed v=s/t=1930.8÷144=[tex] \frac{1930.8}{144} = \frac{160.9}{12} =[/13.408m/s ~nearly]
4 0
3 years ago
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