Can someone please help me with my physics hw

A student produces a power of p = 0.87 kw while pushing a block of mass m = 75 kg on an inclined surface making an angle of Î¸ =

Paul [167]

When block is pushed upwards along the inclined plane

the net force applied on the block will be given as

$F_{net} = mg sin\theta + \mu_k mg cos\theta$

here we know that

m = 75 kg

$\theta = 8.5 degree$

$\mu_k = 0.16$

now plug in all values into this

$F_{net} = 75\times 9.8 sin8.5 + 0.16 \times 75\times 9.8 cos8.5$

$F = 225 N$

now for finding the power is given as

$P = Fv$

$0.87 \times 10^3 = 225 \time v$

$v = \frac{870}{225} = 3.87 m/s$

6 0

3 years ago

The Earth's magnetic field is approximately 0.5 Gauss. What is the magnetic energy density of this field

ser-zykov [4K]

Answer:

The magnetic energy density is $E_B = 9.95 *10^{-4} \ J/m^3$

Explanation:

From the question we are told that

The earths magnetic field is $B = 0.5 \ Gauss$

Now

$1 \ Gauss \to 1 *10^{-4} \ T$

=> $0.5 \ Gauss \to x T$

So

$x = \frac{0.5 * 1*10^{-4}}{1}$

$x = B= 5.0 *10^{-5} \ T$

Generally the magnetic energy density is mathematically represented as

$E_B = \frac{B^2}{ 2 * \mu_o }$

Here $\mu_o$ is the permeability of free space with a constant value

$\mu_o = 4\pi * 10^{-7} N/A^2$

substituting values for equation above

$E_B = \frac{(5*10^{-5})^2}{ 2 * 4\pi * 10^{-7} }$

$E_B = 9.95 *10^{-4} \ J/m^3$

5 0

3 years ago

The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the magnitude of the

aksik [14]

Answer:

Option A

The drag force goes up by a factor of 4

Explanation:

Fact:

Drag force is directly proportional to the square of the velocity of a moving object

Therefore, when drag force pushes opposite to the motion as one rides a bicycle, the magnitude of drag force increases by a factor of 4

3 0

4 years ago

Number 25 and 26 please!!!

skad [1K]

laborious ... enclosed may help ... cos -angle = cos +plus angle ... sin - anfle = - sine angle

cos is even function, sine is odd ... draw rhe graohs

6 0

4 years ago

The uniform dresser has a weight of 91 lb and rests on a tile floor for which μs = 0.25. If the man pushes on it in the horizont

Rudiy27

Answer:

F = 22.75 lb

μ₁ = 0.15

Explanation:

The smallest force required to move the dresser must be equal to the force of friction between the man and the dresser. Therefore,

F = μR

F = μW

where,

F = Smallest force needed to move dresser = ?

μ = coefficient of static friction = 0.25

W = Weight of dresser = 91 lb

Therefore,

F = (0.25)(91 lb)

F = 22.75 lb

Now, for the coefficient of static friction between shoes and floor, we use the same formula but with the mas of the man:

F = μ₁W₁

where,

μ₁ = coefficient of static friction between shoes and floor

W₁ = Weight of man = 151 lb

Therefore,

22.75 lb = μ₁ (151 lb)

μ₁ = 22.75 lb/151 lb

μ₁ = 0.15

7 0

4 years ago