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eimsori [14]
3 years ago
5

Using this formula a = F/m What acceleration results from exerting a 125N force on a 0.65kg

Physics
1 answer:
Softa [21]3 years ago
5 0

Answer:

Acceleration = 192.3 m/s² (Approx.)

Explanation:

Given:

Force = 125 N

Mass of ball = 0.65 kg

Find:

Acceleration

Computation:

We know that;

Acceleration = Force / Mas

So,

Acceleration = 125 / 0.65

Acceleration = 192.3 m/s² (Approx.)

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A bus covers 10 km in 7 minutes. Find the speed of the bus in km/h
Shtirlitz [24]

Answer:

Explanation: so how many minutes are in an hour 60 right, and the bus travels 10km in 7 minutes right so use math the bus travels 14km in 10 minutes so the bus travels 98km in an hour

8 0
3 years ago
Are Carbon and oxygen are both metalloids?
Marizza181 [45]
Just carbon is a metalloid not oxygen
7 0
3 years ago
PLEASE HELP ASAP!! What are 3 types of forces acting on your marshmallow (or other flying object) when it impacts it’s target? C
Alekssandra [29.7K]
•Every action has an equal and opposite reaction (the object is putting force on the target, and the target is putting an equal amount of force back)
•Am object in motion (the object) will stay in motion unless an outside force acts upon it (the Target)

And as for the third one I really don’t know, those seem to be the only two, I’m sorry. I did what a could, Hope it kinda helps :)
8 0
3 years ago
A 44.0 kg uniform rod 4.90 m long is attached to a wall with a hinge at one end. The rod is held in a horizontal position by a w
pychu [463]

Answer:

x ≤ 3.6913 m

Explanation:

Given

Mrod = 44.0 kg

L = 4.90 m

Tmax = 1450 N

Mman = 69 kg

A: left end of the rod

B: right end of the rod

x = distance from the left end to the man

If we take torques around the left end as follows

∑τ = 0   ⇒   - Wrod*(L/2) - Wman*x + T*Sin 30º*L = 0

⇒   - (Mrod*g)*(L/2) - (Mman*g)*x + Tmax*Sin 30º*L = 0

⇒  -  (44*9.8)*(4.9/2) - (69*9.8)*x + (1450)*(0.5)*(4.9) = 0

⇒ x ≤ 3.6913 m

4 0
3 years ago
Two objects are dropped from rest from the same height. Object A falls through a distance during a time t, and object B falls th
UNO [17]

Answer:

Distance covered by B is 4 times distance covered by A

Explanation:

For an object in free fall starting from rest, the distance covered by the object in a time t is

s=\frac{1}{2}gt^2

where

s is the distance covered

g is the acceleration due to gravity

t is the time elapsed

In this problem:

- Object A falls through a distance s_A during a time t, so the distance covered by object A is

s_A=\frac{1}{2}gt^2

- Object B falls through a distance s_B during a time 2t, so the distance covered by object B is

s_B=\frac{1}{2}g(2t)^2 = 4(\frac{1}{2}gt^2)=4s_A

So, the distance covered by object B is 4 times the distance covered by object A.

5 0
3 years ago
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