Answer:
-0.01 mm
Explanation:
We are given that
The value of one division of vernier scale =0.5 mm
The value of one main scale division=0.49 mm
We have to find the value of least count of the instrument in mm.
We know that
Leas count of vernier caliper=1 main scale division-1 vernier scale division
Least count of vernier caliper=0.49-0.50=-0.01 mm
Hence, the least count of the instrument=-0.01 mm
Answer: -0.01 mm
Answer:
3600joules
Explanation:
formula :W=FS
W=work done (J)
F=force (N)
S=displacement moved in the direction of force (m)
200N×18m
=3600J
Answer:
solution:
to find the speed of a jogger use the following relation:
V
=
d
x
/d
t
=
7.5
×m
i
/
h
r
...........................(
1
)
in Above equation in x and t. Separating the variables and integrating,
∫
d
x
/7.5
×=
∫
d
t
+
C
or
−
4.7619
=
t
+
C
Here C =constant of integration.
x
=
0 at t
=
0
, we get: C
=
−
4.7619
now we have the relation to find the position and time for the jogger as:
−
4.7619 =
t
−
4.7619
.
.
.
.
.
.
.
.
.
(
2
)
Here
x is measured in miles and t in hours.
(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),
to get:
= −
4.7619
=
1
−
4.7619
= −
3.7619
or x
=
7.15
m
i
l
e
s
(b) To find the jogger's acceleration in m
i
l
/
differentiate
equation (1) with respect to time.
we have to eliminate x from the equation (1) using equation (2).
Eliminating x we get:
v
=
7.5×
Now differentiating above equation w.r.t time we get:
a
=
d
v/
d
t
=
−
0.675
/
At
t
=
0
the joggers acceleration is :
a
=
−
0.675
m
i
l
/
=
−
4.34
×
f
t
/
(c) required time for the jogger to run 6 miles is obtained by setting
x
=
6 in equation (2). We get:
−
4.7619
(
1
−
(
0.04
×
6 )
)^
7
/
10=
t
−
4.7619
or
t
=
0.832
h
r
s
Answer:
v = 3.27 m/s
Explanation:
KE = 1/2 mv^2
695 J = 1/2 (130kg)(v^2)
695 J / (1/2 x 130kg) = v^2
v^2 = square root of 10.69
v = 3.27 m/s
