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Brums [2.3K]
3 years ago
11

What other experiments can you do with bubble gum?

Physics
1 answer:
Elanso [62]3 years ago
7 0
What happens when you leave it in differnt soda pops.
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A short-wave radio antenna is supported by two guy wires, 155 ft and 175 ft long. Each wire is attached to the top of the antenn
Vladimir79 [104]

Answer:

163.8 ft

Explanation:

In triangle ABD

AB = 155 ft

Cos63 = \frac{BD}{AB} = \frac{BD}{155}\\BD = 155 Cos63 \\BD = 70.4 ft

Sin63 = \frac{AD}{AB} = \frac{AD}{155} \\AD = 166 Sin63\\AD = 148 ft

Using Pythagorean theorem in triangle ADC

AC^{2} = AD^{2} + DC^{2} \\175^{2} = 148^{2} + DC^{2} \\DC = 93.4 ft

d = distance between the anchor points

distance between the anchor points is given as

d = BD + CD = 70.4 + 93.4\\d = 163.8 ft

5 0
3 years ago
Read 2 more answers
Calculate the rate of loss of heat through a glass window of area 200 CM square and thickness 0.5 CM where temperature inside is
saw5 [17]

Answer:

The inner and outer surfaces of a 0.5-cm thick 2-m by 2-m window glass in winter are 10°C and 3°C, respectively. If the thermal conductivity of the glass

Explanation:

7 0
3 years ago
In the reaction 2Ca + O2 → 2CaO for every 2 Ca you will need how much O2?
Serjik [45]

Answer:

1

Explanation: that is the ratio

5 0
2 years ago
The crankshaft in a race car goes from rest to 3000 rpm in 3.0 s . (a) What is the crankshaft's angular acceleration?
Talja [164]

Answer:

75 rotations

Explanation:

f0 = 0, f = 3000 rpm = 50 rps, t = 3 s

(a) use first equation of motion for rotational motion

w = w0 + α t

2 x 3.14 x 50 = 0 + α x 3

α = 104.67 rad/s^2

(b) Let θ be the angular displacement

use second equation of motion for rotational motion

θ = w0 t + 1/2 α t^2

θ = 0 + 0.5 x 104.67 x 3 x 3

θ = 471.015 rad

The angle turn in one rotation is 2 π radian.

Number of rotation = 471.015 / (2 x 3.14) = 75 rotations

7 0
3 years ago
An attacker at the base of a castle wall 3.60 m high throws a rock straight up with speed 8.00 m/s from a height of 1.70 m above
ryzh [129]

Answer:

we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances

Explanation:

given data

base = 3.60 m

speed u = 8 m/s

height = 1.70 m

to find out

check change in speed

solution

we know here formula for v  that is

v² = u² - 2gh      ............1        for upward speed

v² = u² + 2gh     ............2        for projected speed

so here put all value and find v with h = 3.60 - 1.70 = 1.9 m

v² = 8² - 2(9.8) 1.9  = 26.76

v² = 8² + 2(9.8) 1.9   = 101.24

v = 5.173  m/s    ..............3

v = 10.061 m/s   ...................4

so change in speed form 3 and 4 equation

change in speed = v - u = 8 - 5.173  = 2.827 m/s     .................5

change in speed = v - u = 10.061 - 8 = 2.061 m/s     ..................6

so now we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances

6 0
3 years ago
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