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Alona [7]
3 years ago
5

What is the acceleration of a softball if it has a mass of 0.5 kg and hits the cathers glove with a force of 25n

Physics
1 answer:
ioda3 years ago
3 0
Acceleration is found if we have the force and mass. 

With the following equation: F = ma, we can find the missing values. 

F = 25n
M = 0.5 kg
a = ?

a = f/m
a = 25/0.5
a = 50

a = 50 m/s

So, the acceleration is 50 m/s^2 
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What is the linear speed of a point on the equator, due to the earth's rotation?
kvv77 [185]
The equatorial radius of the earth is
r = 6378 km = 6378 x 10³ m

The earth makes 1 revolution in 24 hours.
The angular velocity is
ω = (2π rad)/(24*3600 s) = 7.2722 x 10⁻⁵ rad/s

The tangential velocity (linear velocity) at a point on the equator is
v = rω
   = (6378 x 10³ m)*(7.2722 x 10⁻⁵ rad/s)
   = 463.8 m/s

Answer: 463.8 m/s

8 0
3 years ago
Devonte pushes a wheelbarrow with 830 W of power. How much work is required to get the wheelbarrow across the yard in 11 s? Roun
zaharov [31]
9,100 is the answer if you do the work so i failed for yall you welcome
8 0
2 years ago
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Two streams merge to form a river. One stream has a width of 8.4 m, depth of 3.5 m, and current speed of 2.2 m/s. The other stre
hoa [83]

Answer:

Explanation:

We shall solve this problem on the basis of pinciple that water is incompressible so volume of flow will be equal at every point .

rate of volume flow of one stream

= cross sectional area x velocity

= 8.4 x 3.5 x 2.2 = 64.68 m³ /s

rate of volume flow of other stream

= 6.6 x 3.6 x 2.7

= 64.15 m³ /s

rate of volume flow of rive , if d be its depth

= 11.2 x d x 2.8

= 31.36 d

volume flow of river = Total of volume flow rate of two streams

31.36 d  = 64.15 + 64.68

31.36 d  = 128.83

d = 4.10 m /s .

6 0
2 years ago
A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w
amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁  + m₂u₂ = v(m₁  +  m₂)

10 x 3   + 25 x 0   = v(10  +  25)

30  = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\

the final kinetic energy of the package;

K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J

the fraction of the initial energy lost;

= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71

7 0
3 years ago
A mass of 25kg is moving and is stopped in 1.2 seconds with an acceleration of 1.0 m/s squared what is the force on the mass?
Phantasy [73]
F=ma therefore 25kg*1.0m/s^2=25N force on the mass
6 0
3 years ago
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