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alina1380 [7]
4 years ago
8

Find the force of a 600 kg car experiences accelerating from 10 km/hr to 35 km/hr in 3 seconds?​

Physics
1 answer:
NNADVOKAT [17]4 years ago
3 0
Given,
m=600 kg
u=10 km/hr
v=35 km/hr
t= 3 sec
a=?
F=?

To find a:
u=10 km/hr=5/18*10=25/6=4.16m/s
v=35 km/hr=5/18*35=175/18=9.72m/s
t=3 sec

Now we have,
u=4.16m/s
v=9.72m/s
t=3 sec

To find acceleration:
a=(v-u)/t
a=( 9.72-4.16)/3
a=5.56/3
a=1.85m/s^2

To find F:
m=600 kg
a=1.85 m/s^2

F=ma
F=600*1.85
F=1110N

Therefore, F=1110N


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The Sun radiates energy at a rate of about 4×1026W. Earth is about 150×106km from the Sun.
sergiy2304 [10]

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8 0
1 year ago
Could someone help me with this<br> you will get 20pts
jarptica [38.1K]

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the circulatory system is things like veins which transport blood cells through the body

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5 0
3 years ago
Read 2 more answers
A cylindrical bucket, open at the top, is 28.0 cm high and 11.0 cm in diameter. A circular hole with a cross-sectional area 1.55
svlad2 [7]

Answer:

so height is 0.1283 m

Explanation:

given data

height = 28 cm

diameter = 11 cm

cross-sectional area = 1.55 cm2

water flow rate  =  2.46×10^−4 m3/s

to find out

How high will the water in the bucket rise

solution

we know that here

potential energy = kinetic energy

mgh = 1/2 mv²

multiply both sides by the 2 and we get

2mgh=mv²

solve it we get

√(2gh) = v    ....................1

h = v²/2g   ...............2

and

flow rate = A V

2.46×10^−4 = V 1.55×10^−4

V = 1.5870 m/s

so from 2

h = v²/2g

h = 1.5870²/ 2(9.81)

h = 0.1283 m

so height is 0.1283 m

6 0
4 years ago
calculate the electrical potential at a point P a distance of 1 m from either two to charge of +10 micro coulomb and -5 micro co
Jobisdone [24]

Answer:

a)  V = 45 10³ V, b) U = 4.59 J

Explanation:

a) The electric potential for a series of point charges is

         V = k ∑ \frac{q_i}{r_i}

in this case point P is at a distance of 1 m from each charge, so the point is located perpendicular to the charges at its midpoint

 

         V = k ( \frac{q_1}{r} + \frac{q_2}{r})

         V = 9 10⁹ (10 - 5/ 1) 10⁻⁶

         V = 45 10³ V

b) the potential energy is

           U = k (  \frac{q_1q}{r} + \frac{q_2q}{r} + \frac{q_1q_2}{r_2} )

where r = 1m and r₂ is the distance between the two charges r₂ = 0.10 m

           U = 9 10⁹ (10 2 / 1 - 5 2/1 - 10 5 /0.10) 10⁻¹²

           U = 9 10⁻³  510

           U = 4.59 J

3 0
3 years ago
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