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Natasha_Volkova [10]
2 years ago
5

Two charges with a certain distance apart have an electrostatic force of 400 N acting

Physics
1 answer:
Doss [256]2 years ago
3 0
As the charges’ distance increase, there is a weaker force of attraction between them hence the electrostatic force decreases as distance increases. It increase by 4 (times 4) so the force will decrease by 4 making the answer

=A (400 divided by 4 = 100)
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An object initially at rest experiences an acceleration of 0.281 m/s2 to the South for a time of 5.44 seconds. It then increases
andre [41]

Answer:

12.0 meters

Explanation:

Given:

v₀ = 0 m/s

a₁ = 0.281 m/s²

t₁ = 5.44 s

a₂ = 1.43 m/s²

t₂ = 2.42 s

Find: x

First, find the velocity reached at the end of the first acceleration.

v = at + v₀

v = (0.281 m/s²) (5.44 s) + 0 m/s

v = 1.53 m/s

Next, find the position reached at the end of the first acceleration.

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²

x = 4.16 m

Finally, find the position reached at the end of the second acceleration.

x = x₀ + v₀ t + ½ at²

x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²

x = 12.0 m

5 0
3 years ago
An object takes 5s to reach the ground from a hight of 5m on planet.what is the value of g on planet?
Greeley [361]
T=5s
S=5m
U=0s/m
Acceleration=m/s^2
S=ut+(1/2)(a)(5^2)
A=(25/2)=5
A=5/(25/2)
A=10/25
A=2/5
A=0.4
A=0.4m/s^2.
8 0
3 years ago
Which of the following do cutoffs form in a meandering river
mrs_skeptik [129]

Answer: An oxbow lake forms after there has been deposition of sediment, by the new cutoff channel flowing adjacent to it, at the entrances of the abandoned bend; this seals the bend off from the rest of the river.

8 0
4 years ago
Please help me! I'll give brainliest!!!!
poizon [28]

Answer:

Because they have already made an impact within our atmosphere

6 0
3 years ago
Read 2 more answers
A 10 kg runaway grocery cart runs into a spring with spring constant 250 n/m and compresses it by 60 cm. what was the speed of t
andrew-mc [135]
The initial kinetic energy of the cart is
K= \frac{1}{2} mv^2 (1)
where m is the mass of the cart and v its initial velocity.

Then, the cart hits the spring compressing it. The maximum compression occurs when the cart stops, and at that point the kinetic energy of the cart is zero, so all its initial kinetic energy has been converted into elastic potential energy of the spring:
U= \frac{1}{2}kx^2
where k is the spring constant and x is the spring compression.

For energy conservation, K=U. We can calculate U first: the compression of the spring is x=60 cm=0.60 m, while the spring constant is k=250 N/m, so
U= \frac{1}{2}kx^2= \frac{1}{2}(250 N/m)(0.60 m)^2=45 J

So, the initial kinetic energy of the cart is also 45 J, and from (1) we can find the value of the initial velocity:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 45 J}{10 kg} }  =3 m/s
7 0
3 years ago
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