Mass of block A above is 25-kg. Calculate the force of gravity on block A.

HELP

Sergeu [11.5K]

Answer:

The answer is: To accelerate an object the force applied to the object has to increase.

Explanation:

the acceleration of an object increases with increased force and decreases with increased mass.

3 0

2 years ago

A printer is connected to a 1.0 m cable. if the magnetic force is 9.1 × x10-5 n, and the magnetic field is 1.3 × 10-4 t, what is

Valentin [98]

The magnetic force on a current-carrying wire due to a magnetic field is given by
$F=ILB$
where
I is the current
L the wire length
B the magnetic field strength

In our problem, L=1.0 m, $F=9.1 \cdot 10^{-5} N$ and $B=1.3 \cdot 10^{-4}T$ , so we can re-arrange the formula to find the current in the wire:
$I= \frac{F}{LB}= \frac{9.1 \cdot 10^{-5} N}{(1.0 m)(1.3 \cdot 10^{-4} T)} =0.7 A$

5 0

2 years ago

A 18Ω resistance is cut into three equal parts and connected in parallel. Find the equivalent resistance of the combination.

Andrew [12]

Answer:

2Ω

Explanation:

If a 18Ω resistance is cut into three equal parts each of the resistance will be 18Ω/3 = 6Ω

Equivalent ratio in parallel is expressed as:

1/R = 1/6 + 1/6 + 1/6

1/R = 3/6

Cross multiply

3R = 6

R = 6/3

R = 2Ω

Hence the required equivalent resistance is 2Ω

5 0

2 years ago

A ranger in a national park is driving at 56 km/h when a deer jumps onto the road 65 m ahead of the vehicle. After a reaction ti

vagabundo [1.1K]

Answer:

t = 1.58 s

Explanation:

given,

Speed of ranger, v = 56 km/h

v = 56 x 0.278 = 15.57 m/s

distance, d = 65 m

deceleration,a = 3 m/s²

reaction time = ?

using stopping distance formula

$d = v. t + \dfrac{v^2}{2a}$

$t = \dfrac{d}{v} -\dfrac{v}{2a}$

t is the reaction time

$t = \dfrac{65}{15.57} -\dfrac{15.57}{2\times 3}$

t = 1.58 s

hence, the reaction time of the ranger is equal to 1.58 s.

3 0

2 years ago

You have two square metal plates with side lengths of (6.50 C) cm. You want to make a parallel-plate capacitor that will hold a

gtnhenbr [62]

Answer:

The necessary separation between the two parallel plates is 0.104 mm

Explanation:

Given;

length of each side of the square plate, L = 6.5 cm = 0.065 m

charge on each plate, Q = 12.5 nC

potential difference across the plates, V = 34.8 V

Potential difference across parallel plates is given as;

$V = \frac{Qd}{L^2 \epsilon_o} \\\\d = \frac{V L^2 \epsilon_o}{Q}$

Where;

d is the separation or distance between the two parallel plates;

$d = \frac{VL^2 \epsilon_o}{Q} \\\\d = \frac{34.8*(0.065)^2 *8.854*10^{-12}}{12.5*10^{-9}} \\\\d = 0.000104 \ m\\\\d = 0.104 \ mm$

Therefore, the necessary separation between the two parallel plates is 0.104 mm

6 0

2 years ago