A force of 500 N is exerted on a baseball by the bat for 0.001 s. What is the change in momentum of the baseball?

A vertical spring with a spring constant of 420 N/m is mounted on the floor. From directly above the spring, which is unstrained

mixas84 [53]

Answer:

19.53 cm

Explanation:

The computation of the height is as follows:

Here we applied the conservation of the energy formula

As we know that

P.E of the block = P.E of the spring

m g h = ( 1 ÷ 2) k x^2

where

m = 0.15

g = 9.81

k = 420

x = 0.037

So now put the values to the above formula

(0.15) (9.81) (h) = 1 ÷2 × 420 × (0.037)^2

1.4715 (h) = 0.28749

h = 0.19537 m

= 19.53 cm

3 0

2 years ago

The acceleration of gravity on the moon is about 1.6 m/sec2. An experiment to test gravity compares the time it takes objects to

wlad13 [49]

I really think that the answer A

3 0

2 years ago

Three point charges, two positive and one negative, each having a magnitude of 20 C are placed at the vertices of an equilateral

Daniel [21]

The resultant force on the positive charge is mathematically given as

X=40N

<h3>What is the magnitude of the electrostatic force on the negative charge?</h3>

Question Parameters:

Three-point charges, two positive and one negative, each having a magnitude of 20

Generally, the -ve charge is mathematically given as

$Q+=\sqrt{x^2+x^2+2x.xcos120}\\\\Q+=\sqrt{2x^2+2x*(1/2)}$

Q+=X

Therefore

$x=\frac{Kq1q2}{r2}\\\\x=\frac{9*10^9*20*10^{-6}*20*10^{-6}}{(30*10^-2)^2}$

X=40N

For more information on Force

brainly.com/question/26115859

5 0

1 year ago

An electron that has a velocity with x component 2.6 × 106 m/s and y component 3.2 × 106 m/s moves through a uniform magnetic fi

elena55 [62]

Answer:

a) $\vec F_{B} = 8.766\times 10^{-14}\,T\,k$ , b) $\vec F_{B} = -8.766\times 10^{-14}\,T\,k$

Explanation:

a) The magnetic force experimented by a particle has the following vectorial form:

$\vec F_{B} = q\cdot \vec v \times \vec B$

The charge of the electron is equal to $-1.602\times 10^{-19}\,C$ . Then, cross product can be solved by using determinants:

$\vec F_{B} = \begin{vmatrix}i&j&k\\-4.165\times 10^{-13}\, C\cdot \frac{m}{s} &-5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}$

The magnetic force is:

$\vec F_{B} = 8.766\times 10^{-14}\,T\,k$

b) The charge of the proton is equal to $1.602\times 10^{-19}\,C$ . Then, cross product has the following determinant:

$\vec F_{B} = \begin{vmatrix}i&j&k\\4.165\times 10^{-13}\, C\cdot \frac{m}{s} &5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}$

The magnetic force is:

$\vec F_{B} = -8.766\times 10^{-14}\,T\,k$

8 0

2 years ago

If an impulse of 400 Ns acts on an object for 15s, what is the force of the object?

Ksju [112]

Answer:

J for impulse

t for time

F for force

formula is J=F×t

Explanation:

putting values in eqs after rearranging

we need to find force so

F=J ÷t

F=400÷15

=26.67

=27(rounded off)

27N is the Force applied.

7 0

2 years ago