Answer:
The acceleration of the crate is 
.
Explanation:
Given that,
Force, F = 750 N
Mass of the crate, m = 250 kg
The coefficient of friction is 0.12.
We need to find the acceleration of the crate. The net force acting on the crate is given by :
f is frictional force, 
So, the acceleration of the crate is . Hence, this is the required solution.
Answer:
11.25m
Explanation:
Given parameters:
Initial velocity = 0m/s
Time of running = 3s
Acceleration = 2.5m/s²
Unknown:
Displacement = ?
Solution:
To solve this problem, we apply one of the motion equations.
S = ut + 
at²
u is the initial velocity
t is the time taken
a is the acceleration
S = (0 x 3) + ( x 2.5 x 3²) = 11.25m
Answer: Different vocal tract and harmonics
Explanation:
We are given that a small boy and a grown woman both speak at approximately the same pitch.
To distinguish between the two, we check their harmonics which is different for both of them owing to the different vocal tracts they have. Owing to different vocal tracts, they each produce different harmonic which is actually the multiple of frequency of the wave. Thus, we can make the determination using this.
Answer:
Benzene must be kept away from flames.
Explanation:
its right on edgenuity
Answer:
A. 23.9
B.22.9
C. The levels will be equal
D. Obviously that will be to maintain atmospheric pressure
Explanation:
For mercury the pressure in both tubes at R is same so
P_left = P_right
Thus
=>>>Po + rho_t x g x (5 + R - 1.5) = Po + rho_ m x g x R
rho_t x g x (5 + R - 1.5) = rho_m x g xR
rho_t x (3.5 + R) = rho_m x R
3.5 + R = (rho_m/rho_t) x R
3.5 + R = (13560/867) x R
3.5 + R = 15.64 x R
R x (15.64 - 1) = 3.5
R = 3.5/14.64
= 0.239 m
= 23.9 cm this is for Mercury
ii)water
similarly,
3.5 + R = (rho_w/rho_t) x R
3.5 + R = (1000/867) x R
3.5 + R = 1.153 x R
R X (1.153 - 1) = 3.5
R = 3.5/0.153
= 22.9m for water
