<h3>Question 1</h3>
Answer
option C) velocity
Explanation
acceleration = Δv ÷ Δt
<h3>Question 2</h3>
Answer
option C) m/s²
Explanation
Δv ÷ Δt
= m/s ÷ s
= m/s x 1/s
= m/s²
<h3>Question 3</h3>
Answer
option B) velocity has both direction and speed.
That is why velocity can be negative but speed can not and velocity is rate of change of displacement where as speed is rate of change of distance.
The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
To find the answer, we need to know about the time of flight and range of projectile motion.
<h3>What's the expression of range of a projectile motion?</h3>
- Range = U²× sin(2θ)/g
- U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
- U=√{Range×g/sin(2θ)}
- Here, range= 2.20m, = 36.5°
- U= √{2.20×9.8/sin(73)}
U= √{2.20×9.8/sin(73)} = 22.5m/s
<h3>What's the expression of time of flight in projectile motion?</h3>
- Time of flight= (2×U×sinθ)/g
- So, T= (2×22.5×sin36.5°)/9.8
= 2.73 s
Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
Learn more about the range and time period of projectile motion here:
brainly.com/question/24136952
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Answer:
F = - 3.56*10⁵ N
Explanation:
To attempt this question, we use the formula for the relationship between momentum and the amount of movement.
I = F t = Δp
Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say
v = d / t
t = d / v
Given that
m = 26 g = 26 10⁻³ kg
d = 50 mm = 50 10⁻³ m
t = d/v
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
F t = m v - m v₀
This is so, because the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵
F = - 3.56*10⁵ N
The negative sign is as a result of the force exerted against the bullet
Answer:
F₁ = 4 F₀
Explanation:
The force applied on the string by the ball attached to it, while in circular motion will be equal to the centripetal force. Therefore, at time t₀, the force on ball F₀ is given as:
F₀ = mv₀²/r --------------- equation (1)
where,
F₀ = Force on string at t₀
m = mass of ball
v₀ = speed of ball at t₀
r = radius of circular path
Now, at time t₁:
v₁ = 2v₀
F₁ = mv₁²/r
F₁ = m(2v₀)²/r
F₁ = 4 mv₀²/r
using equation (1):
<u>F₁ = 4 F₀</u>
In a projectile, the horizontal acceleration is zero. The velocity remains constant at all times. However, the <u>vertical acceleration</u> is -9.81m/s^2.
Hope this helps!