The gravitational force exerted by the earth on a person standing on the earth's surface is 602.74 N.
<h3>What is the gravitational force of the earth on the person?</h3>
The gravitational force exerted by the earth on a person standing on the earth's surface is given below as follows:
where
G = 6.67 * 10⁻¹¹
m¹ = 62 kg
m² = 5.97 * 10²⁷ kg
r = 6.4 * 10⁶ m
Therefore, the gravitational force exerted by the earth on a person standing on the earth's surface is 602.74 N.
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Answer:
T=0.372 s, f=2.7 Hz, w=16.9 rad/s, k=179.2 N/m, v= 8.78 m/s, F= 48.4 N
Explanation:
a.)
Period: It is already given in the question "oscillator repeats its motion every 0.372 s".
So T=0.372 s
b)
frequency= f = 1/ T
f = 1/ 0.372
f=2.7 Hz
c).
Angular frequency= w= 2πf
w= 2*π*2.7
w=16.9 rad/s
d)
Spring Constant:
As w=
⇒w²= k/m
⇒k= m*w²
⇒k= 0.628 * 16.9² N/m
⇒k=179.2 N/m
e)
The mass will have maximum speed when it passes through the mean position.
At mean position
Maximum elastic potential energy = Maximum kinetic energy
1/2 k A² = 1/2 m v² ( A is amplitude of oscillation)
⇒ v=
⇒ v= 
\
⇒ v= 8.78 m/s
f)
Maximum force will be exerted on the block when it is at maximum distance.
F= k* A ( A is amplitude of oscillation)
F= 179.2 * 0.27 N
F= 48.4 N
1)
<Solve using the formula which is:
Mass=Density×Volume
2)
3)
(Length × Width × Height)
(Answer=7.5)
Answer:
v = 88.89 [m/s]
Explanation:
To solve this problem we must use the principle of conservation of momentum which tells us that the initial momentum of a body plus the momentum added to that body will be equal to the final momentum of the body.
We must make up the following equation:
where:
F = force applied = 4000 [N]
t = time = 0.001 [s]
m = mass = 0.045 [kg]
v = velocity [m/s]
![4000*0.001=0.045*v\\v=88.89[m/s]](https://tex.z-dn.net/?f=4000%2A0.001%3D0.045%2Av%5C%5Cv%3D88.89%5Bm%2Fs%5D)
