75.17 mg of the radioactive substance will remain after 24 hours.
Answer:
Explanation:
Any radioactive substance will obey the exponential decay behavior. So according to this behavior, any radioactive substance will be decaying in terms of exponential form of disintegration constant and Time.
Disintegration constant is the rate of decay of radioactive elements. It can be measured using the half life time of the radioactive element .While half life time is the time taken by any radioactive element to decay half of its concentration. Like in this case, at first the scientist took 200 mg then after 17 hours, it got reduced to 100 g. So the half life time of this element is 17 hours.
Then Disintegration constant = 0.6932/Half Life time
Disintegration constant = 0.6932/17=0.041
Then as per the law of disintegration constant:
Here N is the amount of radioactive element remaining at time t and 
is the initial amount of sample, x is the disintegration constant.
So here, = 200 mg, x = 0.041 and t = 24 hrs.
N = 200 ×
=75.17 mg.
So 75.17 mg of the radioactive substance will remain after 24 hours.
Answer:
The flux is 682.6 Wb.
Explanation:
Given that,
Vector field 
We need to calculate the flux
Using formula of flux
Put the value into the formula
Hence, The flux is 682.6 Wb.
The cars will have equal speeds and the 2 kg car will have greater kinetic energy.
Answer:
Fa=774 N
Fb=346 N
Explanation:
We will solve this problem by equating forces on each axis.
- On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
- On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative
While towing we know that car is mot moving in y-direction so net force in y-axis must be zero
⇒∑Fy=0
⇒
⇒
⇒
Given that resultant force on car is 950N in positive x-direction
⇒∑Fx=950
⇒
⇒
⇒
⇒
⇒
⇒ 
⇒
Therefore approximately, Fa=774 N and Fb=346 N
Answer:
v₀ₓ = 15 m / s, 
= 5.2 m / s
v = 15.87 m / s
, θ = 19.1
Explanation:
This is a projectile launch problem. The horizontal speed that is constant throughout the entire path is worth 15 m / s, instead the vertical speed changes in value due to the acceleration of gravity, let's look for the initial vertical speed
Vy² =² - 2 g y
² = 
² + 2 g y
= √ (² + 2 gy
Let's calculate
= √ (1.25² + 2 9.8 1.3)
= √ (27.04)
= 5.2 m / s
The initial speed can be calculated by the initial speed
v = √ v₀ₓ² + ²
v = RA (15² + 5.2²)
v = 15.87 m / s
We look for the angle with trigonometry
tan θ = voy / vox
θ = tan⁻¹ I'm going / vox
θ = tan⁻¹ 5.2 / 15
θ = 19.1
The answer is
v₀ₓ = 15 m / s
= 5.2 m / s
