All categories

2 years ago

During cold periods called ____________________, large parts of Earth’s surface were covered with sheets of ice.

Physics

2 answers:

uysha [10]2 years ago

6 0

Ice age i think because it sound like it

Darya [45]2 years ago

6 0

Answer:

During cold periods called 'ice age', large parts of Earth’s surface were covered with sheets of ice.

Explanation:

Ice age refers to the coldest period on earth.

The ice age was the period in which ice glaciers and sheets expand across the earth. That's why it is known as the cold periods.

The complete statement is: During cold periods called ice age, large parts of Earth’s surface were covered with sheets of ice.

You might be interested in

Eric has a mass of 80 kgkg. He is standing on a scale in an elevator that is accelerating downward at 1.7 m/s2m/s2. What is the

kolezko [41]

Answer:

The answer will be 936 N.

Explanation:

Given that

m = 80 kg

Acceleration of the elevator , a= 1.7 m/s² ( upward)

The gravity force on the mass = m g

The reading on the scale = F N

Now by applying the Newton's second law

F - m g = ma

F= m g + m a

F= m ( g +a )

F= 80 ( 10 + 1.7 ) N ( take g= 10 m/s²)

F=80 x 11.7 N

F= 936 N

Therefore the reading on the scale will be 936 N.

The answer will be 936 N.

5 0

3 years ago

A person of mass m will bungee-jump from a bridge over a river where the height from the bridge to the river is h. The bungee co

Likurg_2 [28]

Answer: $k_{min}=\frac{18mg}{h}$

Explanation:

Given

mass of person is m

Distance between bridge and river is h

chord has an un-stretched length of $\frac{2h}{3}$

Let spring constant be k

Person will just stop before hitting the river

Conserve energy i.e. Potential Energy of Person is converted in to elastic energy of chord

$mgh=\frac{kx^2}{2}$

$x=h-\frac{2h}{3}=\frac{h}{3}$

$mgh=\frac{kh^2}{18}$

$k=\frac{18mg}{h}$

Thus $k_{min}=\frac{18mg}{h}$

5 0

3 years ago

What is the value of 9682 when rounded to three significant figures

sdas [7]

9700 because 682 rounds to 700

5 0

2 years ago

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

R = universal gas constant = $\frac{8.314 \ J}{28.97 \ kg.K}$

k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

$c_p= 1.004 \ kJ/kg.K$

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

$P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}$

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

$P_2 = 17.79 \ bar$

Therefore, the exit pressure is 17.79 bar

7 0

2 years ago

A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 18.5 kg, and the

meriva

Answer:

a. $2.86 m/s^2$

b.1058 N

Explanation:

We are given that

Mass of each dog,M=18.5 kg

Mass of sled with rider,m=250 kg

a.Average force,F=185 N

$\mu_s=0.14$

$g=9.8 m/s^2$

By Newton's second law

$8F-f=(8M+m)a$

$a=\frac{8F-f}{8M+m}=\frac{8(185)-(0.14)(9.8)(250)}{8(18.5)+250}$

$a=2.86 m/s^2$

b.By Newton's second law

$T=ma+\mu_s mg$

Substitute the values

$T=250\times 2.86+0.14(250)(9.8)=1058 N$

Hence, the force in the coupling between the dogs and the sled=1058 N

8 0

3 years ago