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3 years ago

What does that mean?

Physics

2 answers:

Veseljchak [2.6K]3 years ago

8 0

Answer:

acceleration

acceleration is the rate at which velocity change

i think

Monica [59]3 years ago

5 0

Answer:

<u>Acceleration</u> is the rate at which the velocity changes .

You might be interested in

Andy Petite pitches a 0.8 kg baseball with a velocity of 67 m/s. Josh Hamilton

kodGreya [7K]

The Impulse delivered to the baseball is 89 kgm/s.

To solve the problem above, we use the formula of impulse.

⇒ Formula:

I = m(v-u)................. Equation 1

Where:

I = Impulse delivered to the baseball
m = mass of the baseball
v = Final velocity of the baseball
u = initial speed of the baseball

From the question,

⇒ Given:

m = 0.8 kg
u = 67 m/s
v = -44 m/s

⇒ Substitute these values into equation 1

I = 0.8(-44-67)
I = 0.8(-111)
I = -88.8
I ≈ -89 kgm/s

Note: The negative tells that the impulse is in the same direction as the final velocity and therefore can be ignored.

Hence, The Impulse delivered to the baseball is 89 kgm/s.

Learn more about impulse here: brainly.com/question/7973509

7 0

2 years ago

What is the approximate weight of the air inside the tire in English Engineering Units (tire outside diameter = 49", rim diamete

Mademuasel [1]

Answer:

1.265 Pounds

Explanation:

Data provided:

Tire outside diameter = 49"

Rim diameter = 22"

Tire width = 19"

Now,

1" = 0.0254 m

thus,

Tire outside radius, r₁ = 49"/2 = 24.5" = 24.5 × 0.0254 = 0.6223 m

Rim radius, r₂ = 22" / 2 = 11" = 0.2794 m

Tire width, d = 19" = 0.4826 m

Now,

Volume of the tire = π ( r₁² - r₂² ) × d

on substituting the values, we get

Volume of air in the tire = π ( 0.6223² - 0.2794² ) × 0.4826 = 0.46877 m³

Also,

Density of air = 1.225 kg/m³

thus,

weight of the air in the tire = Density of air × Volume air in the tire

weight of the air in the tire = 1.225 × 0.46877 = 0.5742 kg

also,

1 kg = 2.204 pounds

Hence,

0.5742 kg = 0.5742 × 2.204 = 1.265 Pounds

3 0

3 years ago

a 16.0 kg cart is being pulled by a 95.4 N force to the right and a 36.0 N force to the left. What is the acceleration of the ca

Inga [223]

Answer:

The cart's acceleration is $\approx 3.71\,\,\frac{m}{s^2}$

Explanation:

Let's start by finding the net force acting on the cart, and then find its acceleration using Newton's 2nd Law.

Net force = 95.4 N -36.0 N = 59.4 N

Now, since we know the cart's mass, we can use Newton's 2nd Law to find the cart's acceleration:

$F=m\,*\,a\\a=\frac{F}{m} \\a=\frac{59.4}{16} \,\,\frac{m}{s^2} \\a\approx 3.71\,\,\frac{m}{s^2}$

8 0

3 years ago

A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

8 0

3 years ago

In ice cap climates, the intense cold _____.

crimeas [40]

Answer:

A. is true

Explanation:

3 0

3 years ago

Read 2 more answers