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3 years ago

5

If you drop something from low height is it faster

1 answer:

JulijaS [17]3 years ago

8 0

Faster to hit the ground because if so yes?

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A 20 kg wagon is rolling to the right across a floor. A person attempts to catch and stop the crate and applies a force of 70 N,

Serggg [28]

Answer:

1.736m/s²

Explanation:

According to Newton's second law;

$\sum F_x = ma_x\\$

$Fm - Ff = ma_x\\$ where;

Fm is the moving force = 70.0N

Ff is the frictional force acting on the body

$Ff = \mu R\\Ff = \mu mg\\$

$\mu$ is the coefficient of friction

m is the mass of the object

g is the acceleration due to gravity

a is the acceleration/deceleration

The equation becomes;

$Fm - Ff = ma_x\\Fm - \mu mg = ma\\$

Substitute the given parameters

$Fm - \mu mg = ma\\70 - 0.18(20)(9.8) = 20a\\70-35.28 = 20a\\34.72 = 20a\\a = \frac{34.72}{20}\\a = 1.736m/s^2\\$

Hence the deceleration rate of the wagon as it is caught is 1.736m/s²

7 0

3 years ago

Which of these is an uneven rhythm movement? Question 5 options:

Musya8 [376]

Answer:

Sliding would be an uneven rhythm because Galloping and skipping has a constant flow of the same movement and same noise.

Explanation:

8 0

2 years ago

Read 2 more answers

Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius

Paha777 [63]

Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

$a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2$

The centripetal acceleration for the second radius; 4.0 m

$a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2$

The centripetal acceleration for the third radius; 6.0 m

$a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2$

The centripetal acceleration for the fourth radius; 8.0 m

$a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2$

The centripetal acceleration for the fifth radius; 10.0 m

$a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2$

6 0

3 years ago

A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

Natasha2012 [34]

Answer:

a) $t=2s$

b) $h_{max}=1024ft$

c) $v_{y}=-256ft/s$

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

$v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2$

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$0=(64ft/s)^2-2(32ft/s^2)(y-960ft)$

$y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft$

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

$h=-16t^2+64t+960$

$1024=-16t^2+64t+960$

$0=-16t^2+64t-64$

Solving the quadratic equation

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$a=-16\\b=64\\c=-64$

$t=2s$

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s$

Since the projectile is going down the velocity at the instant it reaches the ground is:

$v=-256ft/s$

5 0

3 years ago

What does the prefix for kilo- mean

artcher [175]

The word/prefix " $kilo$ " means $one$ $thousad$

Good luck on your assignment and enjoy your day!

~ $MeIsKaitlyn :)$

3 0

3 years ago