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kifflom [539]
3 years ago
9

Sorin hits a cricket ball straight up into the air. After 1.5s the ball is falling straight down with a speed of 6.5m We want to

find the initial vertical velocity of the ball. We can ignore air resistance.
Which kinematic formula would be most useful to solve for the target unknown?
Physics
1 answer:
dmitriy555 [2]3 years ago
7 0

Answer:

v_0 = v(t)+gt

Explanation:

The vertical velocity of an object in free fall is given by:

v(t) = v_0 +at

where

v0 is the initial vertical velocity

a is the acceleration, which actually corresponds to g=-9.8 m/s^2 (gravitational acceleration, with downward direction)

t is the time

So the explicit formula would be

v(t) = v_0 +gt

And we can re-write it to find the initial vertical velocity:

v_0 = v(t)-gt

For our problem, we have

v(t) = -6.5 m/s (with negative sign since the ball is falling downward)

t = 1.5 s

Substituting, we find

v_0 = -6.5 m/s -(-9.8 m/s^2)(1.5 s)=8.2 m/s

in the upward direction.

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Vector A has a magnitude of 50 units and points in the positive x direction. A second vector, B , has a magnitude of 120 units a
Alex Ar [27]

A) Vector A

The x-component of a vector can be found by using the formula

v_x = v cos \theta

where

v is the magnitude of the vector

\theta is the angle between the x-axis and the direction of the vector

- Vector A has a magnitude of 50 units along the positive x-direction, so \theta_A = 0^{\circ}. So its x-component is

A_x = A cos \theta_A = (50) cos 0^{\circ}=50

- Vector B has a magnitude of 120 units and the direction is \theta_B = -70^{\circ} (negative since it is below the x-axis), so the x-component is

B_x = B cos \theta_B = (120) cos (-70^{\circ})=41

So, vector A has the greater x component.

B) Vector B

Instead, the y-component of a vector can be found by using the formula

v_y = v sin \theta

Here we have

- Vector B has a magnitude of 50 units along the positive x-direction, so \theta_A = 0^{\circ}. So its y-component is

A_y = A sin \theta_A = (50) sin 0^{\circ}=0

- Vector B has a magnitude of 120 units and the direction is \theta_B = -70^{\circ}, so the y-component is

B_y = B sin \theta_B = (120) sin (-70^{\circ})=-112.7

where the negative sign means the direction is along negative y:

So, vector B has the greater y component.

8 0
4 years ago
What makes steel durable
sdas [7]

It is durable because it is one of the strongest metals and doesn't corrode easily.

4 0
3 years ago
The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371
arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

q=m\times c\times (T_2-T_1)

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = 4.18J/g.K

T_1 = initial temperature of water = 293.0 K

T_2 = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

q=250g\times 4.18J/g.K\times (371.2-293.0)K

q=81719J

Now we have to calculate the enthalpy of combustion of octane.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

\Delta H=\frac{-81719J}{0.015mole}

\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

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What direction does an applied force move an object
monitta
The object will move in the direction of the applied force.
4 0
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Read 2 more answers
HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!<br> I NEED THE ANSWER NOW
FrozenT [24]
90 F = 43 OR 0.9F = 0.43
(F = 43 / 90 OR 0.43 / 0.9 =) 0.48 N

upwards force = downwards force
(R =) 1.2 N
5 0
3 years ago
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