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alexandr402 [8]

2 years ago

15

A home run is hit in such a way that the baseball just clears a wall 21m hgh located 130m from home plate the ball is hit at an

angle of 35 degrees to the horizontal find the inital sped of the ball the time it takes the ball to reach the wall and the velocity compontets and the speed if the ball when it reached the wall assume the that ball is hit at a height of 1m above the ground?

1 answer:

svlad2 [7]2 years ago

3 0

Answer:

see the attachment

Explanation:

take coordinate system correctly. use formulas of projectile motion

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The kinetic energy of an object with a mass of 6.8 kg and a velocity of 5.0 m/s is J. (Report the answer to two significant figu

abruzzese [7]

Answer:

$\boxed{\sf Kinetic \ energy \ (KE) = 85 \ J}$

Given:

Mass (m) = 6.8 kg

Speed (v) = 5.0 m/s

To Find:

Kinetic energy (KE)

Explanation:

Formula:

$\boxed{ \bold{\sf KE = \frac{1}{2} m {v}^{2} }}$

Substituting values of m & v in the equation:

$\sf \implies KE = \frac{1}{2} \times 6.8 \times {5}^{2}$

$\sf \implies KE = \frac{1}{ \cancel{2}} \times \cancel{2} \times 3.4 \times 25$

$\sf \implies KE =3.4 \times 25$

$\sf \implies KE = 85 \: J$

8 0

3 years ago

Read 2 more answers

Humans have three types of cone cells in their eyes, which are responsible for color vision. Each type absorbs a certain part of

xxMikexx [17]

Answer:

Frequency of the light will be equal to $6.97\times 10^{1}Hz$

Explanation:

We have given wavelength of the light $\lambda =430nm=430\times 10^{-9}m$

Velocity of light is equal to $v=3\times 10^8m/sec$

We have to find the frequency of light

We know that velocity is equal to $v=\lambda f$ , here $\lambda$ is wavelength and f is frequency of light

So frequency of light will be equal to $f=\frac{v}{\lambda }=\frac{3\times 10^8}{430\times 10^{-9}}=6.97\times 10^{1}Hz$

So frequency of the light will be equal to $6.97\times 10^{1}Hz$

5 0

2 years ago

if the pressure on a gas increases, the volume of the gas also increases is that statement true or false?

Arada [10]

False, according to Boyle's law it's pressure increases, volume decreases

6 0

3 years ago

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.330. At what minimum rate wo

Helga [31]

Answer: $3.23 m/s^2$

Explanation:

Given

$\mu_s =0.330$

Frictional Force is balanced by force due to car acceleration

Frictional force $F_s$

$F_s=ma_{min}$

$\mu_sN=ma_{min}$

$\mu_s\cdot mg=ma_{min}$

$a_{min}=\mu_s \cdot g=0.330\times 9.8=3.23 m/s^2$

6 0

2 years ago

The spacing between the plates of a 1.0 μF capacitor is 0.050 mm. a) What is the surface area of the plates? b) How much charge

nataly862011 [7]

Answer:

(a) surface area of the plate will be equal to $1.129m^2$

(b) Charge on the capacitor is equal to $1.5\times 10^{-6}C$

Explanation:

We have given spacing between the plates d = 0.05 mm = $5\times 10^{-5}m$

Value of capacitance $C=1\mu F=10^{-6}F$

(A) Capacitance of a parallel plate capacitor is equal to $C=\frac{\epsilon _0A}{d}$

So $10^{-6}=\frac{8.85\times 10^{-12}\times A}{10^{-5}}$

$A=1.129m^2$

So surface area of the plate will be equal to $1.129m^2$

(B) It is given that capacitor is charged by 1.5 volt

So voltage V = 1.5 volt

Charge on the capacitor is equal to $Q=CV$

So $Q=1.5\times 10^{-6}C$

5 0

2 years ago