2 years ago

The temperature of a body falls from 30°C to 20°C in 5 minutes. The air

Physics

1 answer:

natulia [17]2 years ago

7 0

Answer:

15.88°C I am not 100% sure this is right but I am 98% sure this IS right

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Scott travels north 5 miles, then goes west 3 miles, and then goes south for 2 miles.

Yuki888 [10]

Scott traveled 10 miles

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2 years ago

A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f

Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

$T=\dfrac{2\pi}{\omega}$

${\omega}=\dfrac{2\pi }{0.2}\ rad/s$

${\omega}=31.41\ rad/s$

We know that

${\omega}^2=m\ K$

K=Spring constant

$K=\dfrac{\omega^2}{m}$

$K=\dfrac{31.41^2}{0.12}\ N/m$

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

3 0

3 years ago

which of this is not example of simple machine? A) a crowbar,B)a knife,C)a bicycle,D)a wheelchair ramp

DaniilM [7]

B. Knife HOpe this helps :3

3 0

3 years ago

A bullet fired vertically at a velocity of 36m/s .after 45 the bullet hit the top of a bulid how height is a bulid?

zzz [600]

Answer:

The height of the building is 8,302.5 m

Explanation:

Given;

velocity of the projectile, u = 36 m/s

time of motion, t = 45 s

Let the upward direction of the bullet be negative,

The height of the building is calculated as;

$h = ut - \frac{1}{2} gt^2\\\\h = (36\times 45) - (\frac{1}{2} \times 9.8 \times 45^2)\\\\h = 1620 - 9922.5\\\\h = -8,302.5 \ m\\\\The \ height \ of \ the \ building \ is \ 8,302.5 \ m$

3 0

2 years ago

A light, flexible cable is wrapped around a solid cylinder with mass 3.3 kg and a radius of 0.8 meters. The cylinder rotates on

kari74 [83]

Answer:

$9.16rad/s^2$

Explanation:

We are given that

Mass, $m_1=3.3 kg$

Radius,r=0.8 m

$m_2=4.9 kg$

Height,h=2.9 m

We have to find the angular acceleration of the cylinder.

According to question

$4.9g-T=4.9a$

$Tr=I\alpha$

Where

$\alpha=\frac{a}{r}$

$Tr=\frac{1}{2}m_1ra$

$T=\frac{1}{2}m_1a=\frac{1}{2}(3.3)a$

Substitute the value

$4.9g-\frac{1}{2}(3.3a)=4.9a$

$4.9\times 9.8=4.9a+\frac{3.3a}{2}$

Where $g=9.8 m/s^2$

$48.02=a(4.9+1.65)=6.55a$

$a=\frac{48.02}{6.55}=7.33m/s^2$

Angular acceleration, $\alpha=\frac{a}{r}=\frac{7.33}{0.8}=9.16rad/s^2$

7 0

3 years ago