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Artyom0805 [142]
4 years ago
13

A snowball is thrown with an initial x velocity of 7.5 m/s and an initial y velocity of 8.4 m/s . Part A How much time is requir

ed for the snowball to reach its highest point
Physics
1 answer:
elena55 [62]4 years ago
8 0

In order to calculate the time taken by the snowball to reach the highest point in its journey, we need to consider the variables along the y-direction.

Let us list out what we know from the question so that we can decide on the equation to be used.

We know that Initial Y VelocityV_{iy} = 8.4 m/s

Acceleration in the Y direction a_{y} = -9.8 m/s^{2}, since the acceleration due to gravity points in the downward direction.

Final Y Velocity V_{fy} = 0 because at the highest point in its path, an object comes to rest momentarily before falling down.

Time taken t = ?

From the list above, it is easy to see that the equation that best suits our purpose here is V_{fy} = V_{iy} + a_{y}t

Plugging in the numbers, we get 0 = 8.4 - (9.8)t

Solving for t, we get t = 0.857 s

Therefore, the snowball takes 0.86 seconds to reach its highest point.

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A wire is stretched between two posts. Another wire is stretched between two posts that are twice as far apart. The tension in t
Margarita [4]

Answer: 996m/s

Explanation:

Formula for calculating velocity of wave in a stretched string is

V = √T/M where;

V is the velocity of wave

T is tension

M is the mass per unit length of the wire(m/L)

Since the second wire is twice as far apart as the first, it will be L2 = 2L1

Let V1 and V2 be the speed of the shorter and longer wire respectively

V1 = √T/M1... 1

V2 = √T/M2... 2

Since V1 = 249m/s, M1 = m/L1 M2 = m/L2 = m/2L1

The equations will now become

249 = √T/(m/L1) ... 3

V2 = √T/(m/2L1)... 4

From 3,

249² = TL1/m...5

From 4,

V2²= 2TL1/m... 6

Dividing equation 5 by 6 we have;

249²/V2² = TL1/m×m/2TL1

{249/V2}² = 1/2

249/V2 = (1/2)²

249/V2 = 1/4

V2 = 249×4

V2 = 996m/s

Therefore the speed of the wave on the longer wire is 996m/s

3 0
3 years ago
Tyrone, an 85-kg teenager, runs off the end of a horizontal pier and lands on a free-floating 130-kg raft that was initially at
Ghella [55]

Answer:u=4.04 m/s

Explanation:

Given

Mass m=85 kg

mass of Raft M=130 kg

velocity of raft and man  v=1.6 m/s

Let initial speed of Tyrone is u

Conserving Momentum as there is no external Force

mu=(M+m)v

85\times u=(85+130)\cdot 1.6

u=\frac{215}{85}\cdot 1.6

u=2.529\cdot 1.6=4.04 m/s  

4 0
3 years ago
The mechanical advantage of a wheel and axle is the radius of the wheel divided by the radius of the axle. TRUE or FALSE.
viva [34]
True! The mechanical advantage of the wheel and axle is equal to the ratio of the radius of the wheel over the radius of the axle.

4 0
3 years ago
What type of stress can cause an anticline?<br><br> A. Compression<br><br> B. force<br><br> C. shear
Andrei [34K]
The stress that can cause on the anticline , is the Compression 
4 0
3 years ago
Read 2 more answers
two ships leave a port at the same time. The first ship sails on a bearing of 40 degrees at 18 knots and the second at a bearing
Yuliya22 [10]

Answer:

The distance between the ships is 87.84 km.

Explanation:

Given that,

Angle of first ship= 40°

Speed of first ship = 18 knots

Angle of second ship= 130°

Speed of second ship = 26 knots

We need to calculate the resultant velocity

Using cosine rule

v=\sqrt{v_{1}^2+v_{2}^2-2v_{1}v_{2}\cos\theta}

Put the value into the formula

v=\sqrt{18^2+26^2-2\times18\times26\times\cos90}

v=\sqrt{18^2+26^2}

v=\sqrt{324+676}

v=10\sqrt{10}

We need to calculate the distance between the ships

d =v\times t

Put the value into the formula

d=10\sqrt{10}\times1.5\times1.852

d=87.84\ km

Hence, The distance between the ships is 87.84 km.

7 0
3 years ago
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