A car traveling with an initial velocity of 12 m/s accelerates at a constant rate of 2.2 m/s2 for a time of 4 seconds.

There are three types of rocks on Earth. Each rock type forms in a different way. Igneous rocks were the first rocks on Earth. T

zubka84 [21]

Answer:

A

Explanation:

hot rocks bois, gottem

5 0

2 years ago

Read 2 more answers

A garden hose has a radius of 0.0120 m, and water initially comes out at a speed of 2.88m/s. Dasha puts her thumb over the end ,

creativ13 [48]

Answer:

v = 12.4 [m/s]

Explanation:

With the speed and Area information, we can determine the volumetric flow.

$V=v*A\\A=\pi *r^{2}$

where:

r = radius = 0.0120 [m]

v = 2.88 [m/s]

$A=\pi *(0.0120)^{2} \\A=4.523*10^{-4} [m]\\$

Therefore the flow is:

$V=2.88*4.523*10^{-4} \\V=1.302*10^{-3} [m^{3}/s ]$

Despite the fact that you cover the inlet with the finger, the volumetric flow rate is the same.

$v=V/A\\v=1.302*10^{-3} /1.05*10^{-4} \\v=12.4[m/s]$

3 0

3 years ago

A stationary 15 kg object is located in a table near the surface of the earth. The coefficient of static friction between the su

madreJ [45]

maximum static friction acting on the object will be

$F_s = \mu_s mg$

plug in all values

$F_s = 0.40 \times 15 \times 9.8 = 58.8 N$

So here it means that if applied force is less than or equal to 58.8 N then the object will remain stationary as friction can balance the external force upto this limit of external force

So here it is given that applied force is 20 N

so here object will not move due to this force and it will remain at rest always

due to this applied force

6 0

3 years ago

_________ = voltage / resistance <br> a. power <br> b. energy <br> c. charge <br> d. current

Fittoniya [83]

c.charge due to the reaction process between the two

4 0

3 years 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 $0.97622\times10^{5}\ Pa$ .

7 0

3 years ago