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2 years ago

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if a 0.040-kg stone is whirled horizontally on the end of a 60-m string at a speed of 4.4 m/s, what is the centripetal force ? *

0.013 n 4 n 0. 32 n 20.1 n

1 answer:

SSSSS [86.1K]2 years ago

4 0

Answer:

0.013N

Explanation:

$F=\frac{mv^{2} }{r}$

m=0.04, v=4.4m/s, r=60

F=0.013 N

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M=3000km v=25m/s what’s the momentum

valina [46]

<h3>Answer:</h3>

Momentum of the given body will be : 75000 Kg m/s

<h3>Explanation:</h3>

According to Newton's first law of motion, all bodies continue to be in the state of rest or motion unless an external force is applied on the body. We can use this in the case of momentum also

The formula of momentum is given by :

$:\implies \sf\quad \sf \: P = mv$

Here, we are given the mass of the body ( m ) as 3000kg and the velocity of the body ( v ) as 25 m/s. On putting the values in the formula:

$\begin{aligned}&:\implies \sf\quad \sf \: P = mv \\& :\implies \sf\quad \sf \: P = 3000 \times 25 \\ & :\implies \sf\quad \sf \: \boxed{ \sf \: P = 75000kgm {s}^{ - 1} } \end{aligned}$

Momentum is associated with the mass of the moving body and can be defined as the quantity of motion measured as a product of mass and velocity.

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1 year ago

Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm

Burka [1]

Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

Explanation:

Given that,

Atmospheric pressure $P_{atm}= 1.013\times10^{5}\ Pa$

drop height h'= 27.1 mm

Density of mercury $\rho= 13.59 g/cm^3$

We need to calculate the height

Using formula of pressure

$p = \rho g h$

Put the value into the formula

$1.013\times10^{5}=13.59\times10^{3}\times9.8\times h$

$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

$h=0.76\ m$

We need to calculate the new height

$h''=h - h'$

$h''=0.76-27.1\times10^{-3}$

$h''=0.76-0.027$

$h''=0.733\ m$

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

$P=\rho g h$

Put the value into the formula

$P= 13.59\times10^{3}\times9.8\times0.733$

$P=0.97622\times10^{5}\ Pa$

Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

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2 years ago

An automobile with tires that have a radius of 0.260 m travels 76000 km before wearing them out. How many revolutions do the tir

Tomtit [17]

Answer:

Explanation:

circumference of the tyre = 2πr = 2 x 3.14 x 0.26 = 1.6328m

76000km = 76000000m

no of revolutions required

= 76000000/1.6328 = 46546 revolutions.

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2 years ago

Can you help me with this paper please I will give you 20 points!

sweet [91]

1) they are attracting because if you look at the arrows they’re all pointing the same way.

2) if the magnet was turned around they would do the opposite and not attract ( this is called repulsion)

3) magnetic pole

4)magnet

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2 years ago

Read 2 more answers

While strolling downtown on a Saturday afternoon, you stumble across an old car show. As you are walking along an alley toward a

snow_tiger [21]

Answer:

1.44 m/s²

Explanation:

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a = Acceleration

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The acceleration of the car is 1.44 m/s²

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2 years ago