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Lera25 [3.4K]
4 years ago
12

Calculate the impulse of a 0.140 kg baseball that starts at rest on a tee and reaches a final velocity of 20 m/s after being str

uck with a force of 100 N.
Physics
2 answers:
oee [108]4 years ago
5 0

Explanation:

<h2>impulse = change in momentum</h2><h2>F×t =m(v-u)</h2><h2>I = 0.140(20-0)</h2><h2>I = 0.140 × 20</h2><h2>I = 2.8NS</h2>
Nastasia [14]4 years ago
5 0

Answer: 2.8 Ns

Explanation:

Given that,

impulse I = ?

Mass of baseball M = 0.140 kg

Initial velocity (at rest) Vi = 0m/s

final velocity Vf = 20 m/s

force F = 100 N

Recall that impulse is the product of force and the time during which it acts

i.e Impulse = Force x Time

(since impulse is also equal to change in momentum, then

impulse = Ft = M x (Vf - Vi)

So, apply I = M x (Vf - Vi)

I = 0.140kg x (20 m/s - 0 m/s)

I = 0.140kg x (20 m/s)

I = 2.8 Ns

Thus, the impulse of the baseball is 2.8 Ns

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Answer:

Explanation:

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7 0
3 years ago
A 1. 18 kg gold cube hangs at the end of a 4. 00 m long string. Rhogold = 19. 3 × 103 kg/m3; rhomercury = 13. 6 × 103 kg/m3. Whe
VashaNatasha [74]

When the gold cube is immersed in mercury, the tension in the string in Newtons is 3.142N.

<h3>What is tension?</h3>

Tension is the force acting on the linear object like string, chain or rope due to pulling.

Volume of gold V = mass / density

V = 1.18 /19.3x 10³ =61.1 x 10⁻⁶ m³

Tension in the string after immersing will be

T = [ρ(Gold)  -ρ(Hg)] g. V

T =[ 19.3x 10³ - 13.6 x 10³] x 9.81 x 61.1 x 10⁻⁶

T =3.416 N
Thus, the tension in the string is 3.42 N.

Learn more about tension.

brainly.com/question/4087119

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6 0
3 years ago
The woman dries herself with a towel. She hangs the wet towel in the bathroom to dry. Figure 2 shows two places she could hang t
Darya [45]

tail rail holds the towel from a certain way where the water climbs down

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3 years ago
A 40.0 turn coil of wire of radius 3.0 cm is placed between the poles of an electromagnet. The field increases from 0 to 0.75 T
nadezda [96]

Answer:

Induced emf, V=3.76\times 10^{-4}\ V

Explanation:

We have,

Number of turns in the coil, N = 40

Radius of coil, r = 3 cm = 0.03 m

The field increases from 0 to 0.75 T at a constant rate in a time interval of 225 s.

It is required to find the magnitude of the induced emf in the coil if the field is perpendicular to the plane of the coil. The induced emf is given by :

\epsilon=\dfrac{d\phi}{dt}

\phi=NBA\cos\theta, is magnetic flux

\epsilon=NA\dfrac{dB}{dt}\\\\\epsilon=\dfrac{40\times \pi (0.03)^2\times (0.75-0)}{225}\\\\\epsilon=3.76\times 10^{-4}\ V

So, the magnitude of induced emf is 3.76\times 10^{-4}\ V.

7 0
3 years ago
A skateboarder, with an initial speed of 2.0 m/s, rolls virtually friction free down a straight incline of length 18 m in 3.3 s.
lesya692 [45]

Initial speed of the skateboarder (u) = 2 m/s

Distance covered (s) = 18 m

Time taken = 3.3 seconds

Let the acceleration be a.

Using seconds equation of motion:

s = ut + \frac{1}{2} a t^2

18 = 2 \times 3.3 + \frac{1}{2}a \times 3.3^2

a = \frac{11.4 \times 2}{10.89}

a = 2.09 m/s^2

Now, Acceleration down the incline = g Sin Θ

g Sin Θ = a

9.8 × Sin Θ = 2.09

Sin Θ = \frac{2.09}{9.8}

Θ = 12.31°

Hence, the angle of the inclined plane is: Θ = 12.31°

4 0
4 years ago
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