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Stells [14]
2 years ago
12

2. FAt=p (remember p=mv). A 2.5Jg baseball moving at 8.5m/s to the left is slowed by a tipped

Physics
1 answer:
TiliK225 [7]2 years ago
3 0

The force of the tripped catch exerted on the 2.5 Kg ball moving at 8.5 m/s to the Left is 160 N

<h3>Data obtained from the question </h3>
  • Initial velocity (u) = 8.5 m/s
  • Final velocity (v) = 7.5 m/s
  • Time (t) = 5 ms = 0.25 s
  • Mass (m) = 2.5 Kg
  • Force (F) = ?

<h3>How to determine the force</h3>

The force exerted on the ball can be obtained as follow:

F = m(v + u) / t

F = [2.5(7.5 + 8.5)]/ 0.25

F = 40 / 0.25

F = 160 N

Thus, the force exerted on the ball is 160 N

Learn more about momentum:

brainly.com/question/250648

#SPJ1

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What is the force when two charged spheres distance is in half​
slamgirl [31]

Answer:

When two spheres, each with charge Q, are positioned a distance Rapart, they are attracted to ... doubled, the electric-force between the two spheres

4 0
3 years ago
A 1200-kg car is travelling east at a rate of 9 m/s. A 1600-kg truck is travelling south at a rate of 13 m/s. The truck accident
weeeeeb [17]

Answer:

Around 3.57m/s

Explanation:

p=mv

Let's denote the momentum, mass, and velocity of the car with the subscript 1, and for the truck use 2. After the collision, the combined momentum can be denoted with the subscript 3.

p_1=1200\cdot 9=10800 \\\\p_2=1600\cdot 13=20800 \\\\20800-10800=10000 \\\\1200kg+1600kg=2800kg \\\\10000=2800v_3 \\\\v_3\approx 3.57m/s

Hope this helps!

3 0
3 years ago
The buoyant force on a submerged boulder acts upward because _______.
vovikov84 [41]
Greater water pressure acts on the bottom than on the top
4 0
3 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
A parallel circuit has a 125 Volt battery connected with 3 resistors. R1= 20 Ω, R2= 100 Ω, and R3= 50 Ω. Find the total current
Musya8 [376]

Answer:

10A

Explanation:

to calculate r

1/R=1/20+1/100+1/50=2/25

r=25/2=12.5

I=125/12.5=10A

8 0
3 years ago
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