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3 years ago

What is the acceleration of a 1.5 kg football thrown with a force of 13.00 N to the

Physics

1 answer:

nignag [31]3 years ago

5 0

Answer:

The acceleration of football is 6.00 m/s².

Explanation:

The acceleration can be calculated using the following equation:

$\Sigma F = ma$

Where:

F: is the force

m: is the mass

a: is the acceleration

$F_{e} - F_{w} = ma$

$13.00 N - 4.00 N = 1.5 kg*a$

$a = \frac{9.00 N}{1.5 kg} = 6.00 m/s^{2}$

Therefore, the acceleration of football is 6.00 m/s².

I hope it helps you!

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What types of electromagnetic radiation does the sun emit?

Ksju [112]

Answer:

https://gml.noaa.gov/education/info_activities/pdfs/LA_radiation.pdf

Explanation:

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2 years ago

The rocket's acceleration has components $a_{x}(t)= \alpha t^{2}$ and $a_{y}(t)= \beta - \gamma t$, where \(\alpha = 2.50 {\

lbvjy [14]

it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]

5 0

3 years ago

PLS SHOW WORK !!!!!!! ASAP

Lynna [10]

Answer:

1- For the track B. The potential energy is the same for the two cars, but because of the slope of the track, the car B earn kinetic energy faster. The gravitation acceleration of the cars will be g•sinθ, and the angle of the track B will have a bigger value for sinθ

2- The conservation of energy applies because the roller coaster is a closed track. When a car climb the track, it earn GPE, which is given by mgh, when it get down in the track, it transform GPE in KE, which is given in 1/2mv².

Position of car (m) GPE KE GPE + KE

top (30m) 60000 0 60000

bottom (0m) 0 60000 60000

halfway down (15m) 30000 30000 60000

three-quarters way down 15000 45000 60000

4 0

2 years ago

The length of aluminum rods produced by a company are approximated by a Gaussian distribution with a mean of 10 cm and a standar

ExtremeBDS [4]

Given Information:

Mean length of aluminum rods = μ = 10 cm

Standard deviation of length of aluminum rods = σ = 0.02 cm

Required Information:

a) P(9.98 < X < 10.02) = ?

b) P(9.90 < X < 10.1) = ?

Answer:

a) P(9.98 < X < 10.02) = 68.27%

b) P(9.90 < X < 10.1) = 100%

Explanation:

What is Normal Distribution?

Normal Distribution or also known as Gaussian Distribution, is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability

a) We want to find out the probability that the length of aluminum rods is between 9.98 and 10.02 cm.

$P(9.98 < X < 10.02) = P( \frac{x - \mu}{\sigma} < Z < \frac{x - \mu}{\sigma} )\\\\P(9.98 < X < 10.02) = P( \frac{9.98- 10}{0.02} < Z < \frac{10.02 - 10}{0.02} )\\\\P(9.98 < X < 10.02) = P( \frac{-0.02}{0.02} < Z < \frac{0.02}{0.02} )\\\\P(9.98 < X < 10.02) = P( -1 < Z < 1 )\\\\$

The z-score corresponding to -1 is 0.15866 and 1 is 0.84134

$P(9.98 < X < 10.02) = P( Z < 1 ) - P( Z < -1 ) \\\\P(9.98 < X < 10.02) = 0.84134 - 0.15866 \\\\P(9.98 < X < 10.02) = 0.6827\\\\P(9.98 < X < 10.02) = 68.27 \%$

Therefore, the probability that the length of aluminum rods is between 9.98 and 10.02 cm is 68.27%

b) We want to find out the probability that the length of aluminum rods is between 9.90 and 10.1 cm.

$P(9.90 < X < 10.1) = P( \frac{x - \mu}{\sigma} < Z < \frac{x - \mu}{\sigma} )\\\\P(9.90 < X < 10.1) = P( \frac{9.90- 10}{0.02} < Z < \frac{10.1 - 10}{0.02} )\\\\P(9.90 < X < 10.1) = P( \frac{-0.1}{0.02} < Z < \frac{0.1}{0.02} )\\\\P(9.90 < X < 10.1) = P( -5 < Z < 5 )\\\\$

The z-score corresponding to -5 is 0 and 5 is 1

$P(9.90 < X < 10.1) = P( Z < 5 ) - P( Z < -5 ) \\\\P(9.90 < X < 10.1) = 1 - 0 \\\\P(9.90 < X < 10.1) = 1\\\\P(9.90 < X < 10.1) = 100 \%$

Therefore, the probability that the length of aluminum rods is between 9.90 and 10.1 cm is 100%

How to use z-table?

Step 1:

In the z-table, find the two-digit number on the left side corresponding to your z-score. (e.g 1.0, 2.2, 0.5 etc.)

Step 2:

Then look up at the top of z-table to find the remaining decimal point in the range of 0.00 to 0.09. (e.g. if you are looking for 1.00 then go for 0.00 column)

Step 3:

Finally, find the corresponding probability from the z-table at the intersection of step 1 and step 2.

8 0

3 years ago

Can someone pls help with the last one

Leviafan [203]

Answer:

ummmmmmmmmmmm ask your teacher or perant for help

Explanation:

so umm ask an adult or teen figure and maybe you will get it write not sure do

6 0

2 years ago