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4 years ago

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you throw a ball vertically so it leaves the ground withe velocity of 3.71m/s. what is its acceleration at this point

1 answer:

tino4ka555 [31]4 years ago

4 0

No matter what direction you throw it, or with what speed, its acceleration is immediately 9.8 m/s^2 downward as soon as you release it from your hand, and it doesn't change until the ball hits something.

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During the first 14 minutes of a 1.0 hour trip, a car has an average speed 36 km/h. What must the average speed in km/h of the c

oksian1 [2.3K]

Answer:

85.5 km/h

Explanation:

$t_{1}$ = time interval for first phase = 14 min = $\frac{14}{60}$ h = 0.233 h

$t_{2}$ = time interval for second phase = 46 min = $\frac{46}{60}$ h = 0.767 h

$v$ = average speed for the entire trip = 74 km/h

$v_{1}$ = average speed in first phase = 36 km/h

$v_{2}$ = average speed in second phase

$d_{1}$ = distance traveled in first phase

$d_{2}$ = distance traveled in first phase

average speed is given as

$v = \frac{d_{1} + d_{2}}{t_{1} + t_{2}}$

$v = \frac{v_{1} t_{1} + v_{2} t_{2}}{t_{1} + t_{2}}$

$74 = \frac{(36) (0.233) + v_{2} (0.767)}{0.233 + 0.767}$

$74 = (36) (0.233) + v_{2} (0.767)$

$v_{2} = 85.5$ km/h

4 0

4 years ago

What does this circuit do? T is the clock All flip flops are D-Flip-Flop What is the function of the circuit if Dser is ‘1’ alwa

Harlamova29_29 [7]

Answer:

That is true

Explanation:

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3 years ago

gayle cooks a roast in her microwave oven. the klystron tube in the oven emits photons whose energy is 1.20 x 10^-3 ev. what are

AveGali [126]

Answer:

$\lambda=1.03\times 10^{-3}\ m$

Explanation:

Given that,

The energy of the microwave oven is $1.2\times 10^{-3}\ eV$ .

We need to find the wavelength of these photons.

$1.2\times 10^{-3}\ eV=1.2\times 10^{-3}\times 1.6\times 10^{-19}\\\\=1.92\times 10^{-22}\ J$

The energy of a wave is given by :

$E=\dfrac{hc}{\lambda}\\\\\lambda=\dfrac{hc}{E}$

Put all the values,

$\lambda=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{1.92\times 10^{-22}}\\\\\lambda=1.03\times 10^{-3}\ m$

So, the wavelength of these photon is $1.03\times 10^{-3}\ m$ .

7 0

3 years ago

A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (

Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants ( $v_{oy}$ = 0) zero, so let's use the equation

y = $v_{oy}$ t -1/2 g t²

y= - ½ g t²

t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

x = vox t

x = v₀ₓ √2y/g

c) Speeds before touching the ground

vₓ = vox = constant

$v_{y}$ = $v_{oy}$ - gt

$v_{y}$ = 0 - g √2y/g

$v_{y}$ = - √2gy

tan θ = Vy / vx

θ = tan⁻¹ (vy / vx)

θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

6 0

3 years ago

PLEASE HELP WITH THIS QUESTION WILL GIVE BRAINLIST TO BEST ANSWER

lys-0071 [83]

The net force is 12 N to the left.

4 0

2 years ago

Read 2 more answers