Answer:
<h3>The answer is 45 J</h3>
Explanation:
The work done by an object can be found by using the formula
<h3>workdone = force × distance</h3>
From the question
distance = 3 meters
force = 15 newtons
We have
workdone = 15 × 3
We have the final answer as
<h3>45 J</h3>
Hope this helps you
Acceleration can be defined as the rate of change in the velocity of an object. Option C is correct.
<h3>What is
Acceleration?</h3>
- It is defined as the rate of change in velocity.
- It can also be defined as the rate of change in position in a particular direction.
Where,
- acceleration
- change in velocity
- time
Therefore, acceleration can be defined as the rate of change in the velocity of an object.
Learn more about Velocity:
brainly.com/question/2239252
Answer:
<h2>
0.147136N/m²</h2>
Explanation:
Pressure is defines as force exerts by a body per its unit area.
Pressure = Force/Area
Given the total force exerted by the gas = 6.05N
Area of the rectangular container = 0.121 m * 0.201 m = 0.024321m²
Pressure of the sample = 6.05/0.02432
Pressure of the sample = 0.147136N/m²
Answer:
The pressure exerted by the woman on the floor is 1.9061 x 10⁷ N/m²
Explanation:
Given;
mass of the woman, m = 55 kg
diameter of the circular heel, d = 6.0 mm
radius of the heel, r = 3.0 mm = 0.003 m
Cross-sectional area of the heel is given by;
A = πr²
A = π(0.003)²
A = 2.8278 x 10⁻⁵ m²
The weight of the woman is given by;
W = mg
W = 55 x 9.8
W = 539 N
The pressure exerted by the woman on the floor is given by;
P = F / A
P = W / A
P = 539 / (2.8278 x 10⁻⁵ )
P = 1.9061 x 10⁷ N/m²
Therefore, the pressure exerted by the woman on the floor is 1.9061 x 10⁷ N/m²
Answer:
xf = 5.68 × 10³ m
yf = 8.57 × 10³ m
Explanation:
given data
vi = 290 m/s
θ = 57.0°
t = 36.0 s
solution
firsa we get here origin (0,0) to where the shell is launched
xi = 0 yi = 0
xf = ? yf = ?
vxi = vicosθ vyi = visinθ
ax = 0 ay = −9.8 m/s
now we solve x motion: that is
xf = xi + vxi × t + 0.5 × ax × t² ............1
simplfy it we get
xf = 0 + vicosθ × t + 0
put here value and we get
xf = 0 + (290 m/s) cos(57) (36.0 s)
xf = 5.68 × 10³ m
and
now we solve for y motion: that is
yf = yi + vyi × t + 0.5 × ay × t
² ............2
put here value and we get
yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s) ²
yf = 8.57 × 10³ m
