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madam [21]
3 years ago
6

a sprinter starts from rest and accelerates at a rate of 0.16 ms ^-2 over a distance of 50.0 meters . How fast is the athlete tr

aveling at the end of the 50.0 meters?
Physics
1 answer:
vlada-n [284]3 years ago
7 0

Answer:

Final velocity v = 4 m/s

Explanation:

Given:

Initial velocity u = 0 m/s

Acceleration a = 0.16 m/s²

Distance s = 50 m

Find:

Final velocity v

Computation:

Using; v² = u² + 2as

v² = (0)² + 2(0.16)(50)

v² = 16 m/s

v = 4 m/s

Final velocity v = 4 m/s

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A crate is pulled with a force of 165 N at an angle 30 ° northwest. What is the resultant horizontal force on the crate?
Ad libitum [116K]

Answer:

Resultant horizontal force = 143 N

Explanation:

Since the a gle is 30° northwest, then it means the resultant force will be horizontal and as such;

Resultant horizontal force = 165 * cos 30

Resultant horizontal force = 142.89

Approximating to a whole number gives;

Resultant horizontal force = 143 N

5 0
1 year ago
Hydrogen is a nonmetal with many nonmetal properties. Why is it on the periodic table with the metals in Group 1?
mrs_skeptik [129]

Answer:

This is because it has ns1 electron configuration like the alkali metals.

Pls Mark Brainiest, am trying to become Virtuoso

6 0
2 years ago
An eagle flying at 35 m/s emits a cry whose frequency is 440 Hz. A blackbird is moving in the same direction as the eagle at 10
Akimi4 [234]

Answer:

a)  F=475.7Hz

b)  F'=410.899Hz

Explanation:

From the question we are told that:

Velocity of eagle V_1=35m/s

Frequency of eagle F_1=440Hz

Velocity of Black bird V_2=10m/s

Speed of sound s=343m/s

a)

Generally the equation for Frequency is mathematically given by

 F=f_0(\frac{v-v_2}{v-v_1})

 F=440(\frac{343-10}{343-35})

 F=475.7Hz

b)

Generally the equation for Frequency is mathematically given by

 F'=f_0(\frac{v+v_2}{v+v_1})

 F'=440(\frac{343+10}{343+35})

 F'=410.899Hz

7 0
3 years ago
Which of the following is a result of gravitational forces in the Solar System?
postnew [5]
The correct answer is d
3 0
3 years ago
Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a dow
Inessa [10]

The magnitude of the current in wire 3  is 2.4 A and in a direction pointing in the downward direction.

  • The force per unit length between two parallel thin current-carrying I_1 and I_2  wires at distance ' r ' is given by  f=\frac{u_0I_1I_2}{2\pi r}   ....(1) .
  • If the current is flowing in both wires in the same direction, and  the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.

A schematic of the information provided in the question can be seen in the image attached below.

From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3

F_2_1=F_2_3

Using equation (1) , we get

\frac{u_0I_2I_1}{0.2} =\frac{u_0I_2I_3}{0.32} \\\\\frac{I_1}{0.2} =\frac{I_3}{0.32} \\\\\frac{1.50}{0.2} =\frac{I_3}{0.32} \\\\0.48=0.2I_3\\\\I_3=2.4A

I₃ = 2.4 A and the current is pointing in the downward direction

Learn more about the magnitude and direction of forces here:

brainly.com/question/14879801?referrer=searchResults

#SPJ4

5 0
1 year ago
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