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3 years ago

The scalar product of two vectors 2N,4N is 4 angle between them

Physics

1 answer:

Angelina_Jolie [31]3 years ago

7 0

Answer:

60°

Explanation:

The scalar product of two vectors with an angle ∅ between them is defined as:

$\overrightarrow{a}\cdot \overrightarrow{b}=|a||b|cos\phi$

Solving for angle ∅:

$\phi= cos^{-1}(\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|a||b|})=cos^{-1}(\frac{4}{8})$

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You throw a glob of putty straight up toward the ceiling, which is 3.50 mm above the point where the putty leaves your hand. The

LuckyWell [14K]

Answer: $4.65\ m/s$

Explanation:

Given

Distance putty has to travel is 3.5 m

The initial speed of putty is 9.50 m/s

Using equation of motion to determine the velocity of putty just before it hits ceiling

$v^2-u^2=2as$

$\Rightarrow v^2-(9.5)^2=2(-9.8)(3.5)\\\\\Rightarrow v^2=9.5^2-68.6\\\Rightarrow v=\sqrt{90.25-68.6}\\\Rightarrow v=4.65\ m/s$

So, the velocity of putty just before hitting is $4.65\ m/s$

5 0

3 years ago

What force is required to accelerate a body with a mass of 15kilograms at a rate of

Paladinen [302]

The force required is

(15 kg) x (the acceleration, in m/s²) newtons.

4 0

3 years ago

Can someone tell me the answer and the reason

spin [16.1K]

Answer:

(a) the blocks all had different masses.

Explanation:

The surface is smooth, therefore coefficient of friction is tending to zero.

Forces for each blocks varied from 6N to 8N to 7N to 5N

The blocks were made of different materials and different materials are going to have varying weight for the same size of block.

8 0

4 years ago

What is the formula for potassium iodide

musickatia [10]

Potassium iodide

Molar mass: 166.0028 g/mol

Formula: KI

3 0

3 years ago

Read 2 more answers

A typical cell phone charger is rated to transfer a maximum of 1.0 Coulomb of charge per second. Calculate the maximum number of

Alexandra [31]

Answer:

The maximum no. of electrons- $2.25\times 10^{22}$

Solution:

As per the question:

Maximum rate of transfer of charge, I = 1.0 C/s

Time, t = 1.0 h = 3600 s

Rate of transfer of charge is current, I

Also,

$I = \frac{Q}{t}$

Q = ne

where

n = no. of electrons

Q = charge in coulomb

I = current

Thus

Q = It

Thus the charge flow in 1. 0 h:

$Q = 1.0\times 3600 = 3600\ C$

Maximum number of electrons, n is given by:

$n = \frac{Q}{e}$

where

e = charge on an electron = $1.6\times 10^{- 19}\ C$

Thus

$n = \frac{3600}{1.6\times 10^{- 19}} = 2.25\times 10^{22}$

3 0

3 years ago