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swat32
2 years ago
5

A minimum temperature thermometer in a standard weather instrument shelter measures an overnight low temperature of 36 degrees F

ahrenheit. Explain why frost was observed on the ground beneath the shelter, even though the recorded minimum temperature was above the freezing point.
Physics
1 answer:
Rudiy272 years ago
5 0

Answer:

Explanation:

Even though the minimum temperature is more than the freezing point, Frost was observed on the ground because the ground will cool rapidly as cool air tends to move towards the ground, the temperature of the ground is lower than the atmosphere a few feet above it.

As the thermometer is kept some feet above the ground so ground temperature may be lower than the minimum recorded temperature.

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A boy jumps into an indoor swimming pool. He notices that the water appears to get colder as he
Viefleur [7K]

Answer:

cold air is more dense than warm water so it sinks to the bottom of the pool

7 0
2 years ago
A 2200 kg truck has put its front bumper against the rear bumper of a 2400 kg SUV to give it a push. With the engine at full pow
leonid [27]

Answer:

a) The maximum possible acceleration the truck can give the SUV is 7.5 meters per second squared

b) The force of the SUV's bumper on the truck's bumper is 18000 newtons

Explanation:

a) By Newton's second law we can find the relation between force and acceleration of the SUV:

F=ma

With F the maximum force the truck applies to the SUV, m the mass of the SUV and a the acceleration of the SUV; solving for a:

a=\frac{F}{m}=\frac{18000}{2400}\approx7.5\,\frac{m}{s^{2}}

b) Because at this acceleration the truck's bumper makes a force of 18000 N on the SUV’s bumper by Third Newton’s law the force of the SUV’s bumper on the truck’s bumper is 18000 N too because they are action-reaction force pairs.

7 0
2 years ago
A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.
fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

F=\dfrac{mv^2}{r}-mg

F=m(\dfrac{v^2}{r}-g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

F=\dfrac{mv^2}{r}+mg

F=m(\dfrac{v^2}{r}+g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}

T=\dfrac{m}{r}(g(r-h)+v^2)

Since, v^2=u^2-2gh

T=\dfrac{m}{r}(u^2-3gh+gr)

Hence, this is the required solution.

6 0
2 years ago
A person weighing 0.70 kn rides in an elevator that has an upward acceleration of 1.5 m/s2. what is the magnitude of the normal
creativ13 [48]
First of all, we can find the mass of the person, since we know his weight W:
W=mg=0.70 kN=700 N
And so
m= \frac{W}{g}= \frac{700 N}{9.81 m/s^2}=71.4 kg

We know for Newton's second law that the resultant of the forces acting on the person must be equal to the product between the mass and the acceleration a of the person itself:
\sum F =  ma
There are only two forces acting on the person: his weight W (downward) and the vincular reaction Rv of the floor against the body (upward). So we can rewrite the previous equation as
R_v -W = ma
We know the acceleration of the system, a=1.5 m/s^2 (upward, so with same sign of Rv), so we can solve to find the value of Rv, the normal force exerted by the elevator's floor on the person:
R_v = ma+W=(71.4 kg)(1.5 m/s^2)+700 N =807N
8 0
2 years ago
Write a hypothesis about the effect of the fan speed on the acceleration of the cart. Use the "if . . . then . . . because . . .
iragen [17]
As an object accelerates i.e., change it's velocity(either direction or speed), the position of the object depends on two factor; If the acceleration was direction based then it might have a zero displacement for eg: if it travels in circle. or it might have a net displacement if it travels in a straight line, quantitatively
s = ut +  \frac{1}{2} at {}^{2}
where,
s = displacement
u = initial velocity
v = final velocity
a = acceleration
t = time

Now, for the hypothesis;
There is no direct relationship between fan speed and acceleration but anyways generally speaking if we do have a relationship that with more fan speed we have a larger displacement of air i.e., a more force i.e., greater acceleration
Thus, it can be said, well not exactly scientific, that with a greater fan speed there will be greater acceleration. if fan speed is increased then acceleration will be more.
:)
8 0
2 years ago
Read 2 more answers
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