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3 years ago

15

How can you remove energy from matter? Question 10 options: By increasing its volume By lowering its temperature By increasing i

ts pressure By boiling it.

1 answer:

Doss [256]3 years ago

5 0

Answer: YA i got u!!! got 20/20, so its correct

Explanation:

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A vertical scale on a spring balance reads from 0 to 250 N. The scale has a length of 15.0 cm from the 0 to 250 N reading. A fis

lesantik [10]

Answer:

Mass of fish=6.252 kg

Explanation:

Force F=0 N to 250 N

Length x=15.0 cm=0.15 m

Frequency f=2.60 Hz

To find

Mass of fish m

Solution

First we need to find spring constant

$k=Force/distance\\k=F/x\\k=250/0.15\\k=1666.7 N/m$

As we know that the time period is given as

$T=2\pi\sqrt{\frac{m}{k} }\\ And\\f_{Frequency} =1/T\\So\\1/f=2\pi\sqrt{\frac{m}{k} }\\1/f^{2}=4\pi^{2}\frac{m}{k}\\ k/f^{2}=4\pi^{2}m\\m=\frac{k}{4\pi^{2}f^{2} } \\m=\frac{1666.7}{4\pi^{2} (2.60)^{2} }\\ m=6.252kg$

4 0

3 years ago

Agora elimine os parênteses dessas expressões

Reil [10]

Answer:

wurk wurk dont stop yes maam or yes sirrrrr

Explanation:

periodt cheat then sis

3 0

4 years ago

(NEED HELP PLEASE) A physics student goes to the roof of the school, 24.15 m above the ground, and drops a pumpkin straight down

slavikrds [6]

Answer:

t = 2.2 s

Explanation:

Given that,

Height of the roof, h = 24.15 m

The initial velocity of the pumpkin, u = 0

We need to find the time taken for the pumpkin to hit the ground. Let the time be t. Using second equation of kinematics to find it as follows :

$h=ut+\dfrac{1}{2}at^2$

Here, u = 0 and a = g

$h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 24.15}{9.8}} \\\\t=2.22\ s$

So, it will take 2.22 s for the pumpkin to hit the ground.

7 0

3 years ago

Using newtons law How much force is accelerated a 66kg skier 2m/s2

Evgen [1.6K]

Answer:

SUPONIENDO QUE NO HAY FRICCIÓN

Explanation:

F = 66 kg( $2 m/s^{2}$ ) = 132 N

8 0

3 years ago

59. (II) The crate shown in Fig. 4-60 lies on a plane tilted at an angle A = 25.0° to the horizontal, with Mk 0.19. (a) Determin

Daniel [21]

Explanation:

a) We need to write down first Newton's 2nd law as applied to the given system. The equations of motion for the x- and y-axes can be written as follows:

$x:\;\;\;\;\;mg\sin 25° - \mu_kN = ma\;\;\;\;\;\;(1)$

$y:\;\;\;\;\;N - mg\cos 25° = 0\;\;\;\;\;\;\;\;\;(2)$

From Eqn(2), we see that

$N = mg\cos 25°\;\;\;\;\;\;\;(3)$

so using Eqn(3) on Eqn(1), we get

$mg\sin 25° - \mu_kmg\cos 25° = ma$

Solving for the acceleration, we see that

$a = g(\sin 25° - \mu_k\cos 25°)$

$\;\;\;\;= 2.45\:\text{m/s}^2$

b) Now that we have the acceleration, we can now solve for the velocity of the crate at the bottom of the plane. Using the equation

$v^2 = v_0^2 + 2ax$

Since the crate started from rest, $v_0 = 0.$ Thus our equation reduces to

$v^2 = 2ax \Rightarrow v = \sqrt{2ax}$

$v = \sqrt{2(2.45\:\text{m/s}^2)(8.15\:\text{m})}$

$\;\;\;\;= 6.32\:\text{m/s}$

6 0

3 years ago