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3 years ago

A force of 8480 N is applied to a cart or decelerate is at a rate of 32.0 m/s2. What is the mass of the cart?

Physics

1 answer:

Otrada [13]3 years ago

4 0

I believe it to be 265kg.

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Which of the following is the newton used to measure? answer

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The newton is described in units of force!

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3 years ago

Two identical speakers are 3.50 m

DiKsa [7]

Answer:

The difference in distance from the speakers is 5.2 - 3.5 = 1.7 m

The listener would be 1/2 wavelength out of phase with the speakers

1/2 y = 1.7 m where y is the wavelength

y = 3.4 m the required wavelength

f = v / y = 340 m/s / 3.4 m = 100 / sec lowest frequency

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3 years ago

How much force is needed to accelerate a 3kg book in 15m/s

kotegsom [21]

F=M×A
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3 years ago

Find the net force and acceleration. 15 points. Will give brainliest!

gladu [14]

Answer:

$\boxed{F_{net} = 28.7 \ N}$

$\boxed{a = 2.1 \ m/s^2}$

Explanation:

Finding the net force:

Firstly , we'll find force of Friction:

$F_{k} = (micro)_{k}mg$

Where $(micro)_{k}$ is the coefficient of friction and m = 13.6 kg

$F_{k} = (0.16)(13.6)(9.8)\\$

$F_{k} = 21.32 \ N$

Now, Finding the net force:

$F_{net} = F - F_{k}\\F_{net} = 50 - 21.32\\$

$F_{net} = 28.7 \ N$

Finding Acceleration:

$a = \frac{F_{net}}{m}$

$a = \frac{28.7}{13.6}$

$a = 2.1 \ m/s^2$

8 0

3 years ago

A figure skater skates across a rink of length 50 m in 12.1 seconds. a. What is the average speed of the skater? (2 points) b. I

melamori03 [73]

(a) The skater covers a distance of S=50 m in a time of t=12.1 s, so its average speed is the ratio between the distance covered and the time taken:
$v= \frac{S}{t}= \frac{50 m}{12.1 s}=4.13 m/s$

(b) The initial speed of the skater is
$v_i = 4 m/s$
while the final speed is
$v_f = 5.3 m/s$
and the time taken to accelerate to this velocity is t=2 s, so the acceleration of the skater is given by
$a= \frac{v_f - v_i}{t}= \frac{5.3 m/s-4.0 m/s}{2.0 s}=0.65 m/s^2$

(c) The initial speed of the skater is
$v_i = 13.0 m/s$
while the final speed is
$v_f=0$
since she comes to a stop. The distance covered is S=8 m, so we can use the following relationship to find the acceleration of the skater:
$2aS=v_f^2 -v_i^2$
from which we find
$a= \frac{-v_i^2}{2S}= \frac{-(13.0m/s)^2}{2 \cdot 8.0 m}=-10.6 m/s^2$
where the negative sign means it is a deceleration.

4 0

4 years ago