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3 years ago

What kind of weather is signaled by cumulus clouds

1 answer:

5 0

Cumulonimbus clouds are associated by severe weather. They are thunderstorm clouds that bring heavy rain, snow, hail, lightning and even tornadoes.

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I'll give brainiest. <br> Approximate how many times greater Saturn's mass is than Mercury's.

Vsevolod [243]

Mass of the saturn = 5.683 × 10^26 Mass of the mercury = 3.285 × 10^23

4 0

3 years ago

Read 2 more answers

Concept Simulation 5.2 reviews the concepts that are involved in this problem. A car is safely negotiating an unbanked circular

bonufazy [111]

Answer:

$v_W=9.23m/s$

Explanation:

The force of friction change is the pavement is dry or wet so to determine the force of friction:

$F=m*a$

$F_k=m*a_c$

$F_k=u_K*N$

$N=m*g$

$F_k=u_K*m*g=m*a_c$

$u_K*g=a_c$

$a_c=\frac{V^2}{R}$

Dry pavement

$u_{KD}*g=\frac{v_D^2}{R}$

Wet pavement

$u_{KW}*g=\frac{v_W^2}{R}$

$u_{KW}=\frac{1}{3}*u_{KD}$

Solve and reduce the factor so:

$\frac{v_W^2}{v_D^2}=\frac{\frac{1}{3}*u_{KD}}{u_{KD}}$

$v_W^2=v_D^2*\frac{1}{3}$

$v_W=v_D*\frac{1}{\sqrt{3}}=16m/s*\frac{1}{\sqrt{3}}$

$v_W=9.23m/s$

5 0

3 years ago

A baseball is thrown at a speed of 30.5.if a baseball has a mass of 0.182.what is momentum?

Oliga [24]

Answer:

p = 5.55 kg-m/s

Explanation:

Given that,

The mass of baseball, m = 0.182 kg

The speed of a baseball, v = 30.5 m/s

We need to find its momentum. The formula for momentum is given by :

p = mv

So,

p = 0.182 kg × 30.5 m/s

p = 5.55 kg-m/s

So, the required momentum is 5.55 kg-m/s.

3 0

3 years ago

A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20 kn. if the modulus of elasticity of the material of

Scilla [17]

Given:
rod of circular cross section is subjected to uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]

From given above, area of cross section = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=force/area
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)

(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183
= 0.0318 (to three significant figures)

(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm (to three significant figures)

5 0

3 years ago

A 25.0-kg child plays on a swing having support ropes that are 2.20 m long. Her brother pulls her back until the ropes are 42.0°

SCORPION-xisa [38]

(a) 139.7 J

The potential energy of the child at the initial position, measured relative the her potential energy at the bottom of the motion, is

$U=mg\Delta h$

where

m = 25.0 kg is the mass of the child

g = 9.8 m/s^2

$\Delta h$ is the difference in height between the initial position and the bottom position

We are told that the rope is L = 2.20 m long and inclined at 42.0° from the vertical: therefore, $\Delta h$ is given by

$\Delta h = L - L cos \theta =L(1-cos \theta)=(2.20 m)(1-cos 42.0^{\circ})=0.57 m$

So, her potential energy is

$U=(25.0 kg)(9.8 m/s^2)(0.57 m)=139.7 J$

(b) 3.3 m/s

At the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:

$U=K=\frac{1}{2}mv^2$

where

m = 25.0 kg is the mass of the child

v is the speed of the child at the bottom position

Solving the equation for v, we find

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(139.7 J)}{25.0 kg}}=3.3 m/s$

(c) 0

The work done by the tension in the rope is given by:

$W=Td cos \theta$

where

T is the tension

d is the displacement of the child

$\theta$ is the angle between the directions of T and d

In this situation, we have that the tension in the rope, T, is always perpendicular to the displacement of the child, d. Therefore, $\theta=90^{\circ}$ and $cos \theta=0$ , so the work done is zero.

7 0

3 years ago