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3 years ago

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A catcher "gives" with a baseball when catching it. If the baseball exerts a force of 437 N on the glove, so that the glove is d

isplaced 9.0 cm, how much work is done by the ball?

1 answer:

sergiy2304 [10]3 years ago

5 0

437x9
is ur answer. I'm not sure tho hope it helps

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A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m

madam [21]

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂ = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

4 0

3 years ago

What is 60mph (miles per hour) in meters per second? ( A mile is 5280ft)<br> please someone help me

777dan777 [17]

Answer:

60mph=26.8224meters per second

Explanation:

4 0

3 years ago

An electron in the n = 7 level of the hydrogen atom relaxes to a lower energy level, emitting light of 397 nm. what is the value

Dimas [21]

Answer:

$n_f=2$

Explanation:

It is given that,

Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})$

$\dfrac{1}{397\times 10^{-9}\ m}=R(\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})$

R = Rydberg constant, $R=1.097\times 10^7\ m^{-1}$

$\dfrac{1}{397\times 10^{-9}\ m}=1.097\times 10^7\ m^{-1}\times (\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})$

Solving above equation we get the value of final n is,

$n_f=2.04$

or

$n_f=2$

So, it will relax in the n = 2. Hence, this is the required solution.

6 0

3 years ago

Read 2 more answers

Three monkeys A, B, and C weighing 20, 26, and 25 lb, respectively, are climbing up and down the rope suspended from D. At the i

Marina86 [1]

Answer: $2235.2lb-ft/s^2$

Explanation:

Given

Mass of monkey A=20lb

Mass of monkey B=26lb

Mass of monkey C=25lb

acceleration of monkey A= $4.2ft/s^2$

acceleration of monkey B=0

acceleration of monkey C= $-5.4ft/s^2$

Force Due to monkey A $\left ( F_A\right )=20\times 4.2 =84lb-ft/s^2\left ( downwards\right )$

Force Due to monkey A $\left ( F_B\right )=26\times 0 =0lb-ft/s^2$

Force Due to monkey A $\left ( F_C\right )=25\times 5.4 =135 lb-ft/s^2\left ( upward\right )$

In addition to it Weights of monkeys will be acting downwards therefore net Downwards force is balanced by tension

T= $\left ( 20+26+25\right )32.2+84-135=2235.2 lb-ft/s^2$

5 0

3 years ago

A horizontal force of 100 N is used to pull a crate, which weighs 500 N, at constant velocity across a horizontal floor. What is

Ilya [14]

Answer:

0.2

Explanation:

Horizontal force=100N

Weight of crate=500 N

We have to find the coefficient of kinetic friction.

Normal ,N=Weight=500N

Horizontal force, $F_x=\mu_kN$

Where $F_x$ =Horizontal force

N=Normal force

$\mu_k$ =Coefficient of kinetic friction

Substitute the values in the formula

$100=\mu_k(500)$

$\mu_k=\frac{100}{500}=0.2$

Hence, the coefficient of kinetic friction =0.2

8 0

3 years ago