Answer:
2.71 m
Explanation:
Force is the product of mass and acceleration
F=m*a
Work done is the product of force and distance
Work done=F*d
In this case;
F= 35 N
Work done = 95 J
95 =35 * d
95 /35 = d
2.71 m= d
Answer:
2. [B] = [L]/[T] and [C] = [L]/[T]
Explanation:
I assume you mean this:
A = B² + 2B⁴/C²
Since you can't add numbers with different units (for example, you can't add seconds to meters), each term in the sum must have the same units as A.
B² = [L]²/[T]²
B = [L]/[T]
B⁴/C² = [L]²/[T]²
C²/B⁴ = [T]²/[L]²
C² = B⁴ [T]²/[L]²
C² = ([L]/[T])⁴ [T]²/[L]²
C² = [L]²/[T]²
C = [L]/[T]
Notice we ignore the 2 coefficient, which is unitless.
Answer:
20m/s
Explanation:
acceleration=final velocity-initial velocity/time
4.0m/s²=v m/s-0m/s/5.0sec
5.0sec×4.0m/s²=v m/s-0m/s×5.0m/s/5.0m/s
20m/s=v
volume of balloon
= 4/3 T R3
= 4/3 x 3.14 x 6.953
= 1405.47 m3
uplift force
= volume of balloon x density of air x 9.8
= = 1405.47 x 1.29 x 9.8
= 1813.05 x 9.8 N
weight of helium gas
= volume of balloon x density of helium x
9.8
= 1405.47 x .179 x 9.8
= 251.58 x 9.8 N
Weight of other mass = 930 x 9.8 N Total weight acting downwards
= 251.58 x 9.8 +930 x 9.8
= 1181.58 x 9.8 N
If W be extra weight the uplift can balance
1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8
1181.58+W=1813.05
W= 631.47 kg
<h3><u>Question</u><u>:</u></h3>
A racing car is travelling at 70 m/s and accelerates at -14 m/s^2. What would the car’s speed be after 3 s?
<h3><u>Statement:</u></h3>
A racing car is travelling at 70 m/s and accelerates at -14 m/s^2.
<h3><u>Solution</u><u>:</u></h3>
- Initial velocity (u) = 70 m/s
- Acceleration (a) = -14 m/s^2
- Time (t) = 3 s
- Let the velocity of the car after 3 s be v m/s
- By using the formula,
v = u + at, we have
- So, the velocity of the car after 3 s is 28 m/s.
<h3><u>Answer:</u></h3>
The car's speed after 3 s is 28 m/s.
Hope it helps
