Answer:
7212.3 N
Explanation:
F = 7.38 x 10^4 N, v = 36.2 m/s
Let b be the strength of magnetic field and charge on the particle is q.
F = q v B Sin theta
Here theta = 90 degree
7.38 x 10^4 = q x 36.2 x B x Sin 90
q B = 2038.7 .....(1)
Now, theta = 17 degree, v = 12.1 m/s
F = q v B Sin theta
F = 2038.7 x 12.1 x Sin 17 ( q v = 2038.7 from equation (1)
F = 7212.3 N
The sleds speed when the spring returns toits uncompressed length is v = 0.03 m/s.
<u>Explanation</u>:
Given,
force constant = 42 N/cm = 0.42 N/m, mass m = 68 kg, spring x = 0.39 m
The potential energy, U, stored in the spring is
U = 1/2 kx^2
= 1 / 2 
0.42 (0.39)^2
= 0.032 J
All its potential energy has been converted into kinetic energy since it has a uncompressed length.
K = 1/2 mv^2
v = sqrt (2K / m)
= √(( 2 0.032) / 68)
v = 0.03 m/s
.
Don’t forget about the formulas!
P=E/T
E=P•T
T=E/P
P= 45,000J / 90sec
500W is your answer
Answer:
9) This is a case of deceleration
10)-0.8 ms-2
b) acceleration is the change in velocity with time
11)
a) 100 ms-1
b) 100 seconds
12) 10ms-1
13) more information is needed to answer the question
14) - 0.4 ms^-2
15) 0.8 ms^-2
Explanation:
The deceleration is;
v-u/t
v= final velocity
u= initial velocity
t= time taken
20-60/50 =- 40/50= -0.8 ms-2
11)
Since it starts from rest, u=0 hence
v= u + at
v= 10 ×10
v= 100 ms-1
b)
v= u + at but u=0
1000 = 10 t
t= 1000/10
t= 100 seconds
12) since the sprinter must have started from rest, u= 0
v= u + at
v= 5 × 2
v= 10ms-1
14)
v- u/t
10 - 20/ 25
10/25
=- 0.4 ms^-2
15)
a=v-u/t
From rest, u=0
8 - 0/10
a= 8/10
a= 0.8 ms^-2
Answer:
You have to a ruler and use that to out line you diagram it super easy and label what categorys you have and details of what explains that category
Explanation:
