emf generated by the coil is 1.57 V
Explanation:
Given details-
Number of turns of wire- 1000 turns
The diameter of the wire coil- 1 cm
Magnetic field (Initial)= 0.10 T
Magnetic Field (Final)=0.30 T
Time=10 ms
The orientation of the axis of the coil= parallel to the field.
We know that EMF of the coil is mathematically represented as –
E=N(ΔФ/Δt)
Where E= emf generated
ΔФ= change inmagnetic flux
Δt= change in time
N= no of turns*area of the coil
Substituting the values of the above variables
=1000*3.14*0.5*10-4
=.0785
E=0.0785(.2/10*10-3)
=1.57 V
Thus, the emf generated is 1.57 V
Answer:
M = 281.25 lb*ft
Explanation:
Given
W<em>man</em> = 150 lb
Weight per linear foot of the boat: q = 3 lb/ft
L = 15.00 m
M<em>max</em> = ?
Initially, we have to calculate the Buoyant Force per linear foot (due to the water exerts a uniform distributed load upward on the bottom of the boat):
∑ Fy = 0 (+↑) ⇒ q'*L - W - q*L = 0
⇒ q' = (W + q*L) / L
⇒ q' = (150 lb + 3 lb/ft*15 ft) / 15 ft
⇒ q' = 13 lb/ft (+↑)
The free body diagram of the boat is shown in the pic.
Then, we apply the following equation
q(x) = (13 - 3) = 10 (+↑)
V(x) = ∫q(x) dx = ∫10 dx = 10x (0 ≤ x ≤ 7.5)
M(x) = ∫10x dx = 5x² (0 ≤ x ≤ 7.5)
The maximum internal bending moment occurs when x = 7.5 ft
then
M(7.5) = 5(7.5)² = 281.25 lb*ft
Answer:
#include <stdio.h>
typedef struct InventoryTag_struct {
int itemID;
int quantityRemaining;
} InventoryTag;
int main(void) {
InventoryTag redSweater;
redSweater.itemID = 314;
redSweater.quantityRemaining = 500;
/* Your solution goes here */
printf("Inventory ID: %d, Qty: %d\n",redSweater.itemID,redSweater.quantityRemaining);
getchar();
return 0;
}
Explanation:
Answer: I have 2 suggestions: one, maybe, maybe, use some of that metal tape that you get at hardware places, and: two, buy a new pipe (or pipes), two be sure that it won't happen again. Have a good day, and thanks for asking the brain!
Explanation:
Answer:
# Initialize a dictionary with the keys
contestants = {"Darci Lynne":0, "Angelica Hale":0, "Angelina Green":0};
# Repeatedly prompt the user for a contestant name to vote for
while True:
# Prompting user for contestant name
cName = input("Enter contestant name to vote: ");
# Checking for Done
if cName.lower() == "done":
break;
# Checking in dictionary
if cName in contestants.keys():
# Updating vote value
contestants[cName] += 1
# New entry
else:
contestants[cName] = 1
# Printing header
print("\n%-20s %-15s\n" %("Contestant Name", "Votes Casted"))
# Printing results
for contestant in contestants:
print("%-23s %-15d" %(contestant, contestants[contestant]))
