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aev [14]
3 years ago
13

1. A fixed pulley is a machine that increases the effort force.

Physics
1 answer:
IRISSAK [1]3 years ago
6 0

For the specific questions given, these would be the answers:

1) False

2) False

3) True

4) True

5) True

I am hoping that these answers have satisfied your queries and it will be able to help you, thank you and have a nice day.

You might be interested in
a toy car travels from A to B at constant speed 30km/hr and without stopping at B return A at constant speed v. if the average s
adelina 88 [10]

Answer:

Speed BA = 18 km/hr.

Explanation:

Given the following data;

Speed AB = 30km/hr

Speed BA = x km/hr

Average speed = 24 km/hr

To find the value of x;

Average speed = (Speed AB + Speed BA)/2

Substituting into the equation, we have

24 = (30 + x)/2

48 = 30 + x

x = 48 - 30

x = 18 km/hr

4 0
3 years ago
A 10kg box accelerates forward at a rate of 12 m/s^2. What is the force acting on the box?
Allushta [10]

Answer:

120N

Explanation:

Newton's second law formula: F= ma

given that m = 10 kg, a = 12 m/s^2

F = ma = 10 kg * 12 m/s^2 = 120 kgm/s^2 = 120 N

5 0
2 years ago
A sports car is advertised to be able to stop in a distance of 50.0 m from a speed is 99.0km/h. What is its acceleration in m/s2
Shtirlitz [24]
99.0km/h =27.5m/s (this is the initial speed)
The final speed is zero
The distance is 50.0m
Therefore you use the formula:
vfinal²=vinitial²+2ad
a=(vfinal²-vinitial²)/2d
 = (0²-27.5²)/(2x50.0)
 =-7.5625 or in correct sigdigs -7.56m/s²
Hope this helps!
4 0
3 years ago
What is 9414 divided by 18​
Levart [38]
9414/18 is equal to 523
3 0
3 years ago
A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. If the
barxatty [35]

Answer:

2.2 meters

Explanation:

Potential energy, PE created by a charge, q at a radius r from the charge source, Q,  is expressed as:

KE=\frac{kQq}{r}\     \ \ \ \ \ \ ...i

k is Coulomb's constant.

#The electric field,E at radius r is expressed as:

E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii

From i and ii, we have:

KE=Eqr

r=(KE)/Eq

#Substitute actual values in our equation:

r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m

Hence, the distance between the charge and the source of the electric field is 2.2 meters

7 0
2 years ago
Read 2 more answers
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