When a horse pulls a cart..

Which step is usually NOT performed when finding a pulse?

NNADVOKAT [17]

The laST ONE I THINK

4 0

2 years ago

A wooden block is sitting on an inclined plane near the bottom. The student gave the block a flick and it moved up the inclined

sladkih [1.3K]

Answer:

The block didn't slide due to balancing of gravitational force with friction force

Explanation:

When the block was given a flick the force provided an acceleration to it and it moved up the inclined plane. when the block reached top it was expected that it would slide back but it didn't this happened because of the frictional force acting on the bottom the block which was balancing the gravitational force component along the plane and this prevented sliding back of the block.

static friction was balancing mg*sin(theta)

fs = mg*sin(theta)

6 0

2 years ago

A football kicked in front of a goal post at an angle of 45 degree to the ground just clear the top by of the post 3m high. Calc

Firlakuza [10]

Answer:

A. 10.84 m/s

B. 1.56 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

Time (T) taken to hit the ground again =?

A. Determination of the velocity of projection.

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

H = u²Sine²θ / 2g

3 = u²(Sine 45)² / 2 × 9.8

3 = u²(0.7071)² / 19.6

Cross multiply

3 × 19.6 = u²(0.7071)²

58.8 = u²(0.7071)²

Divide both side by (0.7071)²

u² = 58.8 / (0.7071)²

u² = 117.60

Take the square root of both side

u = √117.60

u = 10.84 m/s

Therefore, the velocity of projection is 10.84 m/s.

B. Determination of the time taken to hit the ground again.

Angle of projection (θ) = 45°

Velocity of projection (u) = 10.84 m/s

Time (T) taken to hit the ground again =?

T = 2uSine θ /g

T = 2 × 10.84 × Sine 45 / 9.8

T = 21.68 × 0.7071 / 9.8

T = 1.56 s

Therefore, the time taken to hit the ground again is 1.56 s.

7 0

2 years ago

Tudy the images about geologic time.

Fittoniya [83]

Answer:

The first flowering plants appeared in the Mesozoic era, not the Paleozoic era

Explanation:

The Mesozoic era is well known and most famous because of the rule of the dinosaurs which were the dominant animals for most of this are. Also, it is the era in which the mammals appeared, though they lived in the shadows of the dinosaurs and only became dominant after their extinction. Another important evolution that took place and is not mentioned very often is the appearance of the first flowering plants. This was a revolutionary trait for the plants, and it helped them to survive in the changing climate on Earth. Soon this trait enabled this type of plants to spread out significantly and to become one of the most dominant organisms on the planet in the following era.

6 0

2 years ago

A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha

USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

$\Phi_E=\frac{Q}{\epsilon_o}$

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

$\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}$

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

$Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C$

Finally, you obtain for E:

$E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}$

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

3 0

2 years ago