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lina2011 [118]
3 years ago
13

A squirrel jumps into the air with a velocity of 7 m/s at an angle of 20 degrees. What is the maximum height reached by the squi

rrel?
Physics
1 answer:
Marina CMI [18]3 years ago
7 0

Answer:

.3m

Explanation:

Apex

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A horse pulls on a wagon. The reaction force is the wagon pulls on the horse.<br><br> true or false?
Katen [24]

Answer:

True

Explanation:

the horse pull on the wagon but friction and the wight + gravity make the wagon pull on the horse ( newton's 2ed law)

6 0
2 years ago
Read 2 more answers
The air in a room has a pressure of 1 atm, a dry-bulb temperature of 24°C, and a wet-bulb temperature of 17°C. Using the psychro
Licemer1 [7]

Answer:

Given that

Dry-bulb temperature(T) =24°C

Wet-bulb temperature(Tw) = 17°C

Pressure ,P = 1 atm

As we know that  psychrometric chart are drawn at constant pressure.

From the diagram

ω= specific humidity

Lets take these two lines   Dry-bulb temperature(T) line and Wet-bulb temperature(Tw) cut at point P

From chart at point P

a)

Specific humidity,ω = 0.00922 kg/kg

b)

The enthalpy ( h)

h=47.59 KJ/kg

c)

The relative humidity, RH

RH= 49.58 %

d)

Specific volume ,

v= 0.853 m³/kg

5 0
3 years ago
A ball is thrown horizontally from the top of a building 21.8 m high. The ball strikes the ground at a point 101 m from the base
riadik2000 [5.3K]

Answer:

t=2.10 s

u= 47.40 m/s

Explanation:

given that

h= 21.8 m

x= 101 m

g=9.8 m/s²

Lets take horizontal speed of  ball = u m/s

The vertical speed of the car at initial condition is zero ( v= 0).

We know that

h=vt+\dfrac{1}{2}gt^2

v= 0 m/s

h=\dfrac{1}{2}gt^2

now by putting the values

21.8 = 1/2 x 9.8 x t²

t=2.10 s

This is time when ball was in motion.

Now in horizontal direction

x = u .t

101 = u x 2.1

u= 47.40 m/s

6 0
3 years ago
a water bomber flying with a horizontal speed of 85m/s at a height of 3000m drops a load on a fire below. How far in front of th
Andreyy89

Answer:

2081.65 m

Explanation:

We'll begin by calculating the time taken for the load to get to the target. This can be obtained as follow:

Height (h) = 3000 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

3000 = ½ × 10 × t²

3000 = 5 × t²

Divide both side by 5

t² = 3000 / 5

t² = 600

Take the square root of both side

t = √600

t = 24.49 s

Finally, we shall determine the distance from the target at which the load should be released. This can be obtained as follow:

Horizontal velocity (u) = 85 m/s

Time (t) = 24.49 s

Horizontal distance (s) =?

s = ut

s = 85 × 24.49

s = 2081.65 m

Thus, the load should be released from 2081.65 m.

3 0
3 years ago
What is the pressure exerted on three floor by a 10.000N crate if the bottom of the crate has the dimensions of 1/2m by 4m
Fantom [35]

Answer:

fine the area then devide force by area

Explanation:

10000/(0.5*4)= 5000 pa

6 0
3 years ago
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