Answer: A Punnett square can be used to predict genotype and phenotypes of offspring from genetic crosses. ... In the P generation, one parent has a dominant yellow phenotype and the genotype YY, and the other parent has the recessive green phenotype and the genotype yy.
Explanation:
Answer:
27.22 m/s
Explanation:
Let the speed of clay before impact is u.
the speed of clay and target is v after impact.
use conservation of momentum
momentum before impact momentum after impact
mass of clay x u = (mass of clay + mass of target) x v
100 x u = (100 + 500) x v
u = 6 v .....(1)
distance, s = 2.1 m
μ = 0.5
final velocity is zero. use third equation of motion
v'² = v² + 2as
0 = v² - 2 x μ x g x s
v² = 2 x 0.5 x 9.8 x 2.1 = 20.58
v = 4.54 m/s
so by equation (1)
u = 6 x 4.54 = 27.22 m/s
thus, the speed of clay before impact is 27.22 m/s.
Answer:
Explanation:
If no one can see it because the lights were out. Did the flea really jump?
What do you want here?
Max height (2.2sin21)²/ 2(9.8) = 3.2 cm
Time of flight 2(2.2sin21)/ (9.8) = 0.16 s
distance of flight (2.2cos21)(0.16) = 33 cm
Answer:
v_f =63 m/s
Explanation:
given,
starting force = 0 N
uniform rate increase to 36 N
time of action of Force = 35 s
mass of the body = 10 Kg
Speed of the object = ?
From the given data
if we plot F-t curve we will get a triangular shape
we know,
Impulse = Area between F-t curve
= (1/2) x base x height
= 0.5 x 35 x 36
= 630 N.s
now use Impulse-momentum theorem
Impulse = change in momentum
630 = 10 x (v_f - vi)
630 = 10 x (v_f - 0)
v_f =63 m/s
Speed of the object at 35 sec is equal to v_f =63 m/s
Answer:
a) t = 0.90 s, b) t = 0.815 s, c) t = 0.90 s, d) x = 3.6 m, e) t = 0.639 s
Explanation:
all these exercises are about kinematics
a) The body is released from rest,
y = y₀ + v₀ t - ½ g t²
in this case when reaching the ground y = 0 and its initial velocity is vo = 0
0 = y₀ + 0 - ½ g t²
t² = 2 y₀ / g
t² = 2 4 /9.81
t² = 0.815
t = √0.815
t = 0.90 s
b) It is thrown upwards at v₀ = 4 m / s
y = y₀ + v₀ t - ½ g t²
in this case the initial and final height is the same
y = y₀ = 0
0 = v₀ t -1/2 g t²
t = 2 v₀ / g
t = 2 4 /9.81
t = 0.815 s
c) the ball is at y₀ = 4 m and its initial velocity is horizontal v₀ = 4 m / s
y = y₀ + v_{oy} t - ½ g t²
0 = y₀ + 0 - ½ g t²
t² = 2 i / g
t² = 2 4 / 9.81
t² = 0.815
t = 0.90 s
d) the horizontal distance traveled is
x = v₀ₓ t
x = 4 0.90
x = 3.6 m
e) We can calculate the time to fall from I = 2 m
y = y₀ + v_{oy} t - ½ g t²
0 = y₀ + 0 - ½ g t²
t² = 2 y₀i / g
t² = 2 2 /9.81
t² = 0.4077
t = 0.639 s
Therefore, when making measurements, you should find readings around this value.
