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3 years ago

14

If velocity is decreasing,then acceleration:

1 answer:

ziro4ka [17]3 years ago

5 0

If velocity is decreasing, then acceleration is in the direction
opposite to the velocity.

If the object is moving in the direction that you call 'positive',
then acceleration is negative.

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The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experi

vlabodo [156]

Answer:

Smallest drop: Water

Largest drop: Dirt

Explanation:

The heat needed to change the temperature of a sample is:

$Q=cm\Delta T$ (1)

with Q the heat (added(+) or removed(-)), c specific heat, m the mass and $\Delta T$ the change in temperature of the sample. So, if we solve (1) for

Sample A:

$\Delta T=-\frac{Q}{cm} =\frac{Q}{4186*4.0}$

$\Delta T=-\frac{Q}{16744}$

Sample B:

$\Delta T=-\frac{Q}{cm} =\frac{Q}{2700*2.0}$

$\Delta T=-\frac{Q}{5400}$

Sample C:

$\Delta T=-\frac{Q}{cm} =\frac{Q}{1050*9.0}$

$\Delta T=-\frac{Q}{9450}$

Note that the numbers 16744, 5400, 9450 are in the denominator of the expression $-\frac{Q}{cm}$ that gives the drop on temperature. so, if Q is the same for the three samples the smallest denominator gives the largest drop and vice versa.

So, the smallest drop is Sample A and the largest is Sample C.

(Important: The minus sign of $\Delta T$ implies the temperature is dropping)

8 0

3 years ago

A ball is thrown directly downward with an initial speed of 7.70 m/s, from a height of 30.2 m. After what time interval does it

KatRina [158]

Answer:

$t = 1.82$

Explanation:

Given

$u = 7.70m/s$ -- initial velocity

$s = 30.2m$ --- height

Required

Determine the time to hit the ground

This will be solved using the following motion equation.

$s = ut + \frac{1}{2}gt^2$

Where

$g = 9,8m/s^2$

So, we have:

$30.2 = 7.70t + \frac{1}{2} * 9.8 * t^2$

$30.2 = 7.70t + 4.9 * t^2$

Subtract 30.2 from both sides

$30.2 -30.2 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9t^2 - 30.2$

$7.70t + 4.9t^2 - 30.2 = 0$

$4.9t^2 + 7.70t - 30.2 = 0$

Solve using quadratic formula:

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

Where

$a = 4.9;\ b = 7.70;\ c = -30.2$

$t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}$

$t = \frac{-7.70\±\sqrt{651.21}}{9.8}$

$t = \frac{-7.70\±25.52}{9.8}$

Split the expression

$t = \frac{-7.70+25.52}{9.8}$ or $t = \frac{-7.70-25.52}{9.8}$

$t = \frac{17.82}{9.8}$ or $t = -\frac{33.22}{9.8}$

Time can't be negative; So, we have:

$t = \frac{17.82}{9.8}$

$t = 1.82$

Hence, the time to hit the ground is 1.82 seconds

7 0

2 years ago

A particle is projected from a point on a horizontal plane and has an initial velocity of 28root 3 m/s at an angle of 60 degree

Lelechka [254]

Answer:

uujjjjjctc7tox7txr9ll8rz8lr5xl8r6l8dl85x8rl5x8rl5x8rl5xrx8l58rk5xr8l5xr6l8xr68lc

4 0

3 years ago

If a 5.00 kg box slides down a ramp inclined at 60.0° above the horizontal, what is the

Basile [38]

Answer:

A 70 kg box is slid along the floor by a 400 n force. The coefficient of friction between the box and the floor is 0. 50 when the box is sliding

8 0

3 years ago

A projectile is fired horizontally from a height of 78.4 m at a speed of 300 m/sec. How far did it travel horizontally before hi

Mice21 [21]

Answer:

Explanation:

Using the formula for calculating range expressed as;

R = U√2H/g

U is the speed = 300m/s

H is the maximum height = 78.4m

g is the acceleration due to gravity = 9.8m/s²

Substitute into the fromula;

R = 300√2(78.4)/9.8

R = 300 √(16)

R = 300*4

R = 1200m

Hence the projectile travelled 1200m before hitting the ground

3 0

3 years ago