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2 months ago

8

A student solving a physics problem for the velocity of an object has applied appropriate physics principles and obtained the ex

pression v=dt−v0, where d=35.0meter, t=9.00second, and v0=3.00meter/second. what is the velocity v?

2 answers:

Maru [3.3K]2 months ago

8 0

Answer : 312 meters per second.

Explanation : The provided formula is v = dt − v0;

Here, v is unknown, while the other values are given;

d equals 35 meters

t equals 9 seconds

and v0 is 3 seconds

By substituting these into the equation, we have,

v = (35 meters × 9 seconds) − 3 seconds

v = 312

Thus, the velocity v equals 312.

Keith_Richards [3.2K]2 months ago

6 0

The velocity v equals 312 m/s

Additional details

Acceleration represents the rate at which velocity changes.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

a = acceleration (m/s²)

v = final velocity (m/s)

u = initial velocity (m/s)

t = time (s)

d = distance (m)

Let's approach this problem systematically!

This task relates to kinematics.

We'll proceed using the following steps:

Given Data:

d = 35.0 m

t = 9.00 s

v0 = 3.00 m/s

Unknown Variable:

v = ?

Solution:

We can find v by plugging the given values into the formula.

$v = dt - v_o$

$v = 35.0( 9.00 ) - 3.00$

$v = 315 - 3$

$v = 312 ~ m/s$

Summary:

Velocity v is 312 m/s for d = 35.0 m, t = 9.00 s, and v0 = 3.00 m/s.

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H = 40 m is the height from which the ball is released.
m = 1 kg is the ball's mass

Assuming g = 9.8 m/s² and disregarding air resistance.

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Read 2 more answers

 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

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Response:

a) The mug makes contact with the ground 0.7425m from the bar's end. b) |V|=5.08m/s θ= -72.82°

Clarification:

To address this issue, we begin with a diagram depicting the situation. (refer to the attached illustration).

a)

The illustration shows that the problem involves motion in two dimensions. To determine how far from the bar the mug lands, we need to find the time the mug remains airborne by examining its vertical motion.

To compute the time, we utilize the following formula with the known values:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

We have $y_{f}=0$ and $v_{y0}=0$ , allowing us to simplify the equation to:

$0=y_{0}+\frac{1}{2}at^{2}$

Now, we can calculate for t:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

We know $y_{0}=1.20m$ and $a=g=-9.8m/s^{2}$

The negative gravity indicates the downward motion of the mug. Hence, we substitute these values into the provided formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

Which results in:

t=0.495s

This time helps us evaluate the horizontal distance the mug traverses. Since:

$V_{x}=\frac{x}{t}$

Solving for x, we have:

$x=V_{x}t$

Substituting the known values yields:

x=(1.5m/s)(0.495s)

This calculates to:

x=0.7425m

b) With the time determining when the mug strikes the ground established, we can find the final velocity in the vertical direction using the formula:

$a=\frac{v_{f}-v_{0}}{t}$

The initial vertical velocity being zero simplifies our calculations:

$a=\frac{v_{f}}{t}$

Thus, we can determine the final velocity:

$V_{yf}=at$

Given that the acceleration equates to gravity (showing a downward effect), we substitute that alongside the previously found time:

$V_{yf}=(-9.8m/s^{2})(0.495s)$

This leads to:

$V_{yf}=-4.851m/s$

Now, we ascertain the velocity components:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

Next, we find the speed by calculating the vector's magnitude:

$|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}$

<pThus:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

Yielding:

|V|=5.08m/s

Lastly, to ascertain the impact direction, we apply the equation:

$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

<pFulfilling this provides:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

<pLeading to:

$\theta = -72.82^{o}$

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