2 years ago

Help pls asap , i do not get it , I’ll name you the brainliest

Physics

1 answer:

34kurt2 years ago

8 0

We know, velocity = Displacement/Time

Therefore, Displacement= Velocity×Time

Here, displacement= 20×5 m = 100 m

Answer: 100 m

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A 0.046 kg golf ball hit by a driver can accelerate from rest to 67 m/s in 1 ms while the driver is in contact with the ball. Ho

Flauer [41]

Answer:

Average force = 67 mn

Explanation:

Given:

Initial velocity u = 0 m/s

Final velocity v = 67 m/s

Time t = 1 ms = 0.001 sec.

Computation:

Using Momentum theory

Change in momentum = F × Δt

(v-u)/t = F × Δt

F × 0.001 = (67 - 0)/0.001

F= 67,000,000

Average force = 67 mn

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2 years ago

A meterstick is placed on a pivot point of 42.5cm and a 45g mass is hung at the 20cm mark. When released the meterstick remains

Vikki [24]

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2 years ago

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A fixed system of charges exerts a force of magnitude 22 N on a 4.0 C charge. The 4.0 C charge is replaced with a 8.0 C charge.

Karo-lina-s [1.5K]

Answer:

44 N

Explanation:

The electrostatic forces between two charges is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is their separation

We notice that the force is directly proportional to the charges.

In this problem, initially we have a force of

F = 22 N

on a q2 = 4.0 C, exerted by a charge q1.

If the charge is doubled,

q2 = 8.0 C

This means that the force will also double, so it will be

$F=22 N \cdot 2 = 44 N$

3 0

2 years ago

If Calcium has 2 valence electrons and Chlorine has 7 valence electrons, how many Cl-1 ions would need to combine with 1 Ca+2 io

Tanya [424]

Answer:

CALCIUM IDENINE ADEININE AND PHOSPHATE

Explanation:

6 0

2 years ago

A cylindrical Nickel rod (9 mm diameter, 50 m long) is pulled in tension with a load of 6,283 N. What would the elongation of th

Norma-Jean [14]

Answer:

0.29 m

Explanation:

9 mm = 0.009 m in diameter

Cross-sectional area $A = \pi d^2/4 = \pi * 0.009^2/4 = 6.36\times 10^{-5} m^2$

Let the tensile modulus of Nickel $E = 170 \times 10^9Pa$ .

The elongation of the rod can be calculated using the following formula:

$\Delta L = \frac{F L}{A E} = \frac{6283*50}{6.36\times 10^{-5} * 170 \times 10^9} = \frac{314150}{1081200} = 0.29 m$

6 0

3 years ago