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sveta [45]
2 years ago
9

A pair of fuzzy dice is hanging by a string from your rearview mirror. while you are accelerating from a stoplight to 28 m/s in

6.0 s, what angle does the string make with the vertical? (assume your acceleration is constant.)
Physics
1 answer:
Lerok [7]2 years ago
4 0
We can use the formula of motion in physics (2nd law od newton) in this problem:
x direction: Fsin ∅ = ma 
y direction: Fcos ∅ -mg = 0
∅ is equal to sin ∅ / cos ∅  or x/y
tan ∅ = ma / mg = a /g

Applying acceleration formula:
v = vo + at ; 28 = 0 + 6a ; a = 4.67 m/s^2
∅ = tan-1 (a/g) = tan-1 (4.67/9.81) = <span>25.4 degrees.</span>
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WHAT IS THE NET FORCE (CENTRIPETAL FORCE) FOR A 1500 KG
worty [1.4K]
A :-) Given - m = 1500 kg
v = 5 m/s
r = 10 m
Solution -
F = mv^2 by r
F = 1500 x 5 x 5 by 10
( cut 5 and 10 because 5 x 2 = 10 )
F = 1500 x 5 by 2
( cut 2 and 1500 because
2 x 750 = 1500 )
F = 750 x 5
F = 3750 N
.:. The centripetal force is 3750 N.

5 0
3 years ago
A 700kg car had 12,000 joules of kinetic energy, Calculate it’s velocity !! please help due right now !!
Margarita [4]

Answer:

velocity of car=5.855 m/s

Explanation:

8 0
3 years ago
A coin completes 18 spins in 12 seconds. The centripetal acceleration of the edge of the coin is 2.2 m/s2. The radius of the coi
stellarik [79]

Answer:

2.5 cm

Explanation:

Number of spins in 12 seconds = 18

Number of spins in 1 second = 18 / 12 = 1.5

Number of spins in one second is called frequency.

f = 1.5 rps

ac = 2.2 m/s^2

Let r be the radius of coin

Centripetal acceleration, ac = r x w^2

2.2 = r x ( 2 x 3.14 x 1.5) ^2

2.2 = r x 88.7364

r = 0.025 m = 2.5 cm

3 0
3 years ago
Read 2 more answers
Blood pressure is usually measured by wrapping a closed air-filled jacket equipped with a pressure gage around the upper arm of
Sever21 [200]

Answer:

a) High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

Explanation:

a) <em>High and low pressures in kilopascals</em>:

101.325 kPa equals 760 mm Hg, then, we can obtain the values by a single conversion:

p_{high} = 120\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}

p_{high} = 15.999\,kPa

p_{low} = 80\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}

p_{low} = 10.666\,kPa

High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) <em>High and low pressures in pounds per square inch</em>:

14.696 psi equals 760 mm Hg, then, we can obtain the values by a single conversion:

p_{high} = 120\,mm\,Hg\times \frac{14.696\,psi}{760\,mm\,Hg}

p_{high} = 2.320\,psi

p_{low} = 80\,mm\,Hg\times\frac{14.696\,psi}{760\,mm\,Hg}

p_{low} = 1.547\,psi

High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) <em>High and low pressures in meter water column in meters water column</em>:

We can calculate the equivalent water column of a mercury column by the following relation:

\frac{h_{w}}{h_{Hg}} = \frac{\rho_{Hg}}{\rho_{w}}

h_{w} = \frac{\rho_{Hg}}{\rho_{w}}\times h_{Hg} (Eq. 1)

Where:

\rho_{w}, \rho_{Hg} - Densities of water and mercury, measured in kilograms per cubic meter.

h_{w}, h_{Hg} - Heights of water and mercury columns, measured in meters.

If we know that \rho_{w} = 1000\,\frac{kg}{m^{3}}, \rho_{Hg} = 13600\,\frac{kg}{m^{3}}, h_{Hg, high} = 0.120\,m and h_{Hg, low} = 0.080\,m, then we get that:

h_{w, high} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.120\,m

h_{w, high} = 1.632\,m

h_{w, low} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.080\,m

h_{w, low} = 1.088\,m

High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

4 0
3 years ago
A differnece in electric potential is required for an electric charge to flow through a wire.
cupoosta [38]

Answer:

TRUE

Explanation:

6 0
2 years ago
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