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A claw hammer is used to pull a nail from a piece of wood. Where should you place your hand on the handle and where should the n

Alex Ar [27]

<span>You will exert less force the more the lever is longer. So, you need to place your hands as high as possible with the hammers handle.</span>

4 0

4 years ago

At an accident scene on a level road, investigators measure a car’s skid mark to be 88 m long. The accident occurred on a rainy

Oduvanchick [21]

Answer:

The the speed of the car is 26.91 m/s.

Explanation:

Given that,

distance d = 88 m

Kinetic friction = 0.42

We need to calculate the the speed of the car

Using the work-energy principle

work done = change in kinetic energy

$W=\Delta K.E$

$\mu\ mg\times d=\dfrac{1}{2}mv^2$

$v^2=2\mu g d$

Put the value into the formula

$v=\sqrt{2\times0.42\times9.8\times88}$

$v=26.91\ m/s$

Hence, The the speed of the car is 26.91 m/s.

3 0

3 years ago

Read 2 more answers

Which three types of electromagnetic waves carry most of the Sun's energy that strikes Earth?

pshichka [43]

Answer:

A.

Explanation:

3 0

4 years ago

Two force A and B at a point at right angles. If their resultant is 50N and their sum is 70N,

kodGreya [7K]

Taking a wild guess here, but it sounds like you're asked to find |A| and |B| given the magnitude of the resultant A + B, i.e. |A + B| = 50 N, and the sum of the individual magnitudes, |A| + |B| = 70 N.

Recall that the dot product of a vector with itself is equal to the square of that vector's magnitude,

A • A = |A|²

Then

|A + B|² = (A + B) • (A + B)

|A + B|² = (A • A) + 2 (A • B) + (B • B)

|A + B|² = |A|² + 2 (A • B) + |B|²

Since A and B are perpendicular to one another, their dot product is

A • B = 0

So it follows that

|A + B|² = |A|² + |B|²

2500 N² = |A|² + |B|²

Substitute |B| = 70 N - |A| and solve for |A| :

2500 N² = |A|² + (70 N - |A|)²

2500 N² = |A|² + 4900 N² - (140 N) |A| + |A|²

2 |A|² - (140 N) |A| + 2400 N² = 0

|A|² - (70 N) |A| + 1200 N² = 0

|A|² - (70 N) |A| = - 1200 N²

|A|² - (70 N) |A| + 1225 N² = - 1200 N² + 1225 N²

(|A| - 35 N)² = 25 N²

|A| - 35 N = ± 5 N

|A| = 35 N ± 5 N

so that |A| = 30 N or |A| = 40 N. If we fix |A| to be one of these, then |B| will have the other value.

So, the magnitudes are |A| = 30 N and |B| = 40 N.

3 0

3 years ago

A lab technician uses laser light with a wavelength of 650 nmnm to test a diffraction grating. When the grating is 42.0 cmcm fro

Bingel [31]

Answer:

221 lines per millimetre

Explanation:

We know that for a diffraction grating, dsinθ =mλ where d = spacing between grating, θ = angle to maximum, m = order of maximum and λ = wavelength of light.

Since the grating is 42.0 cm from the screen and its first order maximum (m = 1) is at 6.09 cm from the center of the pattern,

tanθ = 6.09 cm/42.0 cm = 0.145

From trig ratios, cot²θ + 1 = cosec²θ

cosecθ = √((1/tanθ)² + 1) = √((1/0.145)² + 1) = √48.562 = 6.969

sinθ = 1/cosecθ = 1/6.969 = 0.1435

Also, sinθ = mλ/d at the first-order maximum, m = 1. So

sinθ = (1)λ/d = λ/d

Equating both expressions we have

0.1435 = λ/d

d = λ/0.1435

Now, λ = 650 nm = 650 × 10⁻⁹ m

d = 650 × 10⁻⁹ m/0.1435

d = 4529.62 × 10⁻⁹ m per line

d = 4.52962 × 10⁻⁶ m per line

d = 0.00452962 × 10⁻³ m per line

d = 0.00452962 mm per line

Since d = width of grating/number of lines of grating

Then number of lines per millimetre = 1/grating spacing

= 1/0.00452962

= 220.77 lines per millimetre

≅ 221 lines per millimetre since we can only have a whole number of lines.

6 0

3 years ago