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Ganezh [65]
3 years ago
8

A ball is thrown straight up in the air and just before it lands it is travelling -47.5 m/s. How long was the ball in the air?

Physics
1 answer:
Setler [38]3 years ago
6 0
Use v = u + at
Message me if you need more help
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What is the approximate uncertainty in the area of a circle of radius 4.5×10^4 cm? Express your answer using one significant fig
Igoryamba

Answer:

A=6.36\times 10^5\ m^2

Explanation:

It is given that,

Radius of the circle, r=4.5\times 10^4\ cm=4.5\times 10^2\ m

The area of the circle is given by :

A=\pi r^2

A=\pi (4.5\times 10^2\ m)^2

A=636172.51\ m^2

or

A=6.36\times 10^5\ m^2

As there is no uncertainty given in the radius of the circle. So, the area of the circle is 6.36\times 10^5\ m^2. Hence, this is the required solution.

5 0
3 years ago
How long does it take a car traveling at 50 mph to travel 75 miles? Use one of the following to find the answer.
Anna35 [415]

Answer:

1.5 hours

Explanation:

Using the equation t=d/v to solve for time:

t=75 miles ÷ 50 miles per hour = 1.5 hours

5 0
2 years ago
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According to the Theory of Continental Drift, what did the Earth’s continents look like 255 million years ago? What is the name
dmitriy555 [2]
The name of this landmas is known as <em></em>Pangaea, was a supercontinent that existed during the late Paleozoic and early Mesozoic eras. It formed approximately 300 million years ago and began to break apart after about 100 million years. 

Theres an image of how this supercontinet looked

3 0
3 years ago
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Particle A of charge 2.70 10-4 C is at the origin, particle B of charge -6.36 10-4 C is at (4.00 m, 0), and particle C of charge
Serhud [2]

Answer:

FC vector representation

F_{C} =(19.03i+13.78j)N

Magnitude of FC

F_{C}=23.495N

Vector direction FC

\beta=35.91 degrees: angle that forms FC with the horizontal

Explanation:

Conceptual analysis

Because the particle C is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

The directions of the individual forces exerted by qA and qB on qC are shown in the attached figure; The force (FAC) of qA over qC is repulsive because they have equal signs and the force (FBC) of qB over qC is attractive because they have opposite signs.

The FAC force is up in the positive direction and the FBC force forms an α angle with respect to the x axis.

\alpha = tan^{-1}(\frac{3}{4}) = 36.86 degrees

To calculate the magnitudes of the forces we apply Coulomb's law:

F_{AC} = \frac{k*q_{A}*q_{C}}{r_{AC}^2} Equation (1): Magnitude of the electric force of the charge qA over the charge qC

F_{BC} = \frac{k*q_{B}*q_{C}}{r_{BC}^2} Equation (2) : Magnitude of the electric force of the charge qB over the charge qC

Known data

k=8.99*10^9 \frac{N*m^2}{C^2}

q_{A}=2.70*10^{-4} C

q_{B}=-6.36*10^{-4} C

q_{C}=1.04*10^{-4} C

r_{AC} =3

r_{BC}=\sqrt{4^2+3^2} = 5

Problem development

In the equations (1) and (2) to calculate FAC Y FBC:

F_{AC} =8.99*10^9*(2.70*10^{-4}* 1.04*10^{-4})/(3)^2=28.05N

F_{BC} =8.99*10^9*(6.36*10^{-4}* 1.04*10^{-4})/(5)^2=23.785N

Components of the FBC force at x and y:

F_{BCx}=23.785 *Cos(36.86)=19.03N

F_{BCy}=23.785 *Sin(36.86)=14.27N

Components of the resulting force acting on qC:

F_{Cx} = F_{ACx}+ F_{BCx}=0+19.03=19.03N

F_{Cy} = F_{ACy}+ F_{BCy}=28.05-14.27=13.78N

FC vector representation

F_{C} =(19.03i+13.78j)N

Magnitude of FC

F_{C}= \sqrt{19.03^2+13.78^2} =23.495N

Vector direction FC

\beta = tan^{-1} (\frac{13.78}{19.03})=35.91 degrees: angle that forms FC with the horizontal

7 0
3 years ago
1. A toy car with mass m1 travels to the right on a frictionless track with a speed of 3 m/s. A second toy car with mass m2 trav
Alika [10]

Answer:

v = 3(m1 - 2m2)/(m1 + m2)

Explanation:

Parameters given:

Velocity of first toy car with mass m1, u1 = 3 m/s (taking the right direction as the positive axis)

Velocity of second toy car with mass m2, u2 = -6 m/s (taking the left direction as the negative x axis)

Using conservation of momentum principle:

Total initial momentum = Total final momentum

m1*u1 + m2*u2 = m1*v1 + m2*v2

Since they stick together after collision, they have the same final velocity.

m1*3 + (m2 * -6) = m1*v + m2*v

3m1 - 6m2 = (m1 + m2)v

v = (3m1 - 6m2) / (m1 + m2)

v = 3(m1 - 2m2) / (m1 + m2)

8 0
3 years ago
Read 2 more answers
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